Derivatives(II) | NEB Class 11 Mathematics Notes

Chapter Overview

Welcome, students! This chapter focuses on mastering the derivatives of Trigonometric, Inverse Trigonometric, Exponential, and Logarithmic functions. In calculus, these functions represent non-linear patterns of growth, oscillation, and decay. Differentiating these functions requires combining your previous knowledge of limits with algebraic methods such as trigonometric identities, logarithmic properties, and implicit differentiation techniques.

What this chapter covers

First Principles for Advanced FunctionsThe Chain Rule for Composite ExpressionsImplicit & Parametric DifferentiationLogarithmic Differentiation TechniquesHigher-Order Derivatives & proofs

Learning Objectives

1

Evaluate the derivatives of trigonometric, exponential, and logarithmic functions from First Principles using standard limits

2

Differentiate composite functions by systematically applying the Chain Rule from the outside in

3

Formulate derivatives for implicit and parametric equations

4

Solve complex variable-exponent problems using the Logarithmic Differentiation technique

5

Calculate higher-order derivatives and construct algebraic proofs for differential equations

The Core Mechanics of Differentiation

Key Insight

Every advanced derivative problem can be simplified by identifying its structure. If a variable is in the exponent (like $x^x$), use natural logarithms. If equations contain mixed coordinates, apply implicit methods. If the variables are connected through an external parameter, divide their individual rates of change.

When deriving trigonometric limits from first principles, your mathematical goal is to isolate the identity $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. This foundational limit is the key to proving advanced trigonometric derivatives.

Key Definitions

Derivative

The exact instantaneous rate of change of a function, defined as the limit of the difference quotient as the increment in the independent variable approaches zero.

$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$

Exponential Function

A mathematical function of the form $f(x) = a^x$, where the base $a$ is a positive constant and $a \neq 1$. The natural exponential function uses base $e \approx 2.718$.

Logarithmic Function

The inverse operation of exponentiation. If $y = a^x$, then $\log_a y = x$. The natural logarithm uses base $e$ and is written as $\ln x$.

Implicit Function

A function in which the dependent variable $y$ is not isolated on one side of the equation, but instead remains structurally mixed with the independent variable $x$.

Higher-Order Derivative

The derivative of a derivative. If we differentiate $y$ once, we get the first derivative $y_1$ or $\frac{dy}{dx}$. Differentiating again yields the second derivative $y_2$ or $\frac{d^2y}{dx^2}$.

Periodic vs. Exponential Rates

Advanced functions display unique rates of change. Periodicity causes sine and cosine functions to cycle, while exponential curves grow or decay rapidly.

Trigonometric Differentiation

$$\frac{d}{dx}(\sin x) = \cos x$$

Deals with periodic, oscillating patterns.

Involves utilizing trigonometric identities (like C-D formulas) during limit derivations.

Requires tracking the negative sign, which applies to any function starting with 'co' (cosine, cotangent, cosecant).

The derivative's value cycles between positive, zero, and negative slopes.

VS

Exponential Differentiation

$$\frac{d}{dx}(e^x) = e^x$$

Deals with rapid growth or decay patterns.

The natural exponential function is unique because its derivative is equal to the function itself.

Differentiating non-natural bases requires scaling by a natural log factor: $\frac{d}{dx}(a^x) = a^x \ln a$.

The derivative is always positive for $a > 1$, representing continuous upward growth.

Memory Key: Trigonometric derivatives switch identities (sine becomes cosine). Exponential derivatives retain their identity and only require scaling factor adjustments.

Formula Sheet — Advanced Functions & Derivations

Derivative of Sine

\frac{d}{dx}(\sin x) = \cos x

Proved from first principles using trigonometric limit properties.

1Let us assume our function is $y = \sin x$.

2Now, let us introduce increments. Let $\Delta x$ be the change in $x$, and $\Delta y$ be the corresponding change in $y$. Therefore, we can write:
$$y + \Delta y = \sin(x + \Delta x)$$

3Next, let us subtract the original equation $y$ from both sides to find the change in $y$:
$$\Delta y = \sin(x + \Delta x) - \sin x$$

4To simplify the right-hand side, we can use the trigonometric C-D formula: $\sin C - \sin D = 2 \cos\frac{C+D}{2} \sin\frac{C-D}{2}$. Let us substitute $C = x + \Delta x$ and $D = x$:
$$\Delta y = 2 \cos\left(x + \frac{\Delta x}{2}\right) \sin\left(\frac{\Delta x}{2}\right)$$

5Now, let us divide both sides of our equation by the step size $\Delta x$ to get:
$$\frac{\Delta y}{\Delta x} = \frac{2 \cos\left(x + \frac{\Delta x}{2}\right) \sin\left(\frac{\Delta x}{2}\right)}{\Delta x}$$

6To prepare for evaluating the limit, we can rewrite this expression by moving the 2 to the denominator:
$$\frac{\Delta y}{\Delta x} = \cos\left(x + \frac{\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}}$$

7Let us apply the limit as $\Delta x \to 0$. We know that $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. By setting $\theta = \frac{\Delta x}{2}$, we find:
$$\frac{dy}{dx} = \cos(x + 0) \cdot 1 = \cos x$$

Derivative of Natural Logarithm

\frac{d}{dx}(\ln x) = \frac{1}{x}

Proved from first principles using standard logarithmic limits.

1Let us write our function as $y = \ln x$.

2Let us introduce increments $\Delta x$ and $\Delta y$, giving us:
$$y + \Delta y = \ln(x + \Delta x)$$

3Subtracting $y$ from both sides gives the increment in $y$:
$$\Delta y = \ln(x + \Delta x) - \ln x$$

4Using the logarithmic property $\ln A - \ln B = \ln\left(\frac{A}{B} ight)$, we can simplify the expression:
$$\Delta y = \ln\left(\frac{x + \Delta x}{x}\right) = \ln\left(1 + \frac{\Delta x}{x}\right)$$

5Now, let us divide by $\Delta x$ and set up our limit:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\ln\left(1 + \frac{\Delta x}{x}\right)}{\Delta x}$$

6Let us use a change of variable to resolve this limit. Let $u = \frac{\Delta x}{x}$, which implies $\Delta x = x u$. As $\Delta x \to 0$, we know $u \to 0$ as well. Therefore, we write:
$$\frac{dy}{dx} = \lim_{u \to 0} \frac{\ln(1 + u)}{x u} = \frac{1}{x} \lim_{u \to 0} \frac{\ln(1 + u)}{u}$$

7We know that the standard limit is $\lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1$. Therefore, we find:
$$\frac{dy}{dx} = \frac{1}{x} \cdot 1 = \frac{1}{x}$$

Analyzing Logarithmic and Exponential Derivatives

When differentiating logarithmic or exponential expressions, use this decision pathway to find the correct method.

Is the base of the log/exponent something other than 'e'?

Yes

Change the base to e: use ln(x)/ln(a) for log_a(x) or a^x = e^(x ln a).

No

Continue to check structural limits ↓

Does the variable appear in both the base and the exponent, like x^x?

Yes

Use Logarithmic Differentiation. Take the natural logarithm of both sides, then differentiate implicitly.

No

Continue to check structural limits ↓

Is the function nested, like ln(sin x) or e^(x^2)?

Yes

Use the Chain Rule. Differentiate the outer function, then multiply by the derivative of the inner function.

No

Apply the standard derivative formula directly.

Formulas and Negative Signs

Trick: Remembering negative signs and matched pairs makes solving advanced functions straightforward.

Steps to Remember:

The 'CO' Derivative Rule: Every trigonometric derivative that starts with the letters 'co' (namely, $\cos x$, $\cot x$, and $\csc x$) has a negative sign in its derivative.
Inverse Matching Pairs: The formulas for inverse sine and inverse cosine are identical, except for their sign. Similarly, tangent and cotangent inverse pairs, as well as secant and cosecant inverse pairs, match with opposite signs.
Logarithmic Exponent Drop: When simplifying complex power terms like $y = [f(x)]^{g(x)}$, taking the natural log of both sides allows you to drop the exponent to the front: $\ln y = g(x) \ln[f(x)]$. This converts a difficult power problem into a straightforward Product Rule problem.

Mnemonic: CO trig terms always differentiate to negative results.

Common Mistakes to Avoid

Avoid these common errors to ensure you retain full marks on your exam papers.

Forgetting the Chain Rule multiplier: Students often write $\frac{d}{dx}(\sin 2x) = \cos 2x$. Remember to multiply by the derivative of the inner term ($2$), which yields the correct answer of $2\cos 2x$.
Omitting dy/dx in implicit terms: When differentiating $y^2$ with respect to $x$, do not write just $2y$. You must include the derivative of $y$ with respect to $x$, writing it as $2y \frac{dy}{dx}$.
Assuming base 'e' for all logarithms: The derivative $\frac{d}{dx}(\log x)$ is only $\frac{1}{x}$ if the base is $e$. If the base is $a$, you must apply the base change formula, which gives $\frac{1}{x \ln a}$.

NEB Exam Traps — Common Board Problems

NEB Exam Warning

Keep these targeted warnings in mind to avoid common pitfalls on board exam questions.

First Principles from composite inputs: When asked to differentiate a function like $\cos 3x$ from first principles, do not write the direct rule. You must expand the limit using trigonometric sum and difference identities.
Higher-order differential proofs: Questions that ask you to prove equations like $(1-x^2)y_2 - xy_1 = 0$ are common. A good strategy is to cross-multiply and square the terms of the first derivative before differentiating a second time, which helps avoid complex fractions.
Infinite nested roots: For questions like $y = \sqrt{\tan x + \sqrt{\tan x + \dots}}$, replace the nested tail of the expression with $y$ to form the equation $y^2 = \tan x + y$, and then differentiate implicitly.

Question Recognition Guide

Keyword / Phrase in Question	Use This
The problem asks for the derivative of a trigonometric function 'from first principles'	First Principles: Set up the limit and apply the trigonometric C-D formula to resolve the indeterminate form.
The variable appears as both the base and the exponent, like x^x or (sin x)^(ln x)	Logarithmic Differentiation: Take the natural logarithm of both sides, simplify the exponent, and differentiate implicitly.
The coordinates are mixed, and the question asks you to 'prove a differential equation'	Implicit Differentiation: Differentiate term-by-term and group dy/dx terms together.
The question presents two equations, x = f(t) and y = g(t)	Parametric Differentiation: Calculate dx/dt and dy/dt separately, and find their ratio.
The problem asks for a higher-order derivative, such as the second (y2) or third (y3) derivative	Successive Differentiation: Find the first derivative, simplify the result, and differentiate again.

Trigonometric Rates of Change

This diagram illustrates how the derivative of the sine function ($\sin x$) corresponds to the cosine function ($\cos x$). Notice that at the peak of the sine curve, the slope is zero, which matches the zero-crossing of the cosine curve. This visualizes the rate of change of oscillating functions.

The relationship between the sine curve and its derivative, the cosine curve, showing how the slope of the tangent changes.

Solved Examples

1

Find the derivative of $e^{3x}$ with respect to $x$.

Easy

Solution

Let us solve this using the Chain Rule. First, let us assume that the inner function is $u = 3x$.

Therefore, we can write our original function as:
$$y = e^u$$
We know that the derivative of $e^u$ with respect to $u$ is $e^u$. Let us write down the derivative of the inner function as well:
$$\frac{du}{dx} = \frac{d}{dx}(3x) = 3$$
The Chain Rule states that:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
Therefore, by substituting our values into this formula, we get:
$$\frac{dy}{dx} = e^{3x} \cdot 3 = 3e^{3x}$$

∴ 3e^(3x)

2

Find $\frac{dy}{dx}$ if $x^2 + y^2 = \sin(xy)$.

Medium

Solution

Since the variables $x$ and $y$ are mixed together, let us perform implicit differentiation.

First, let us differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\sin(xy))$$
We know that the derivative of $x^2$ is $2x$, and the derivative of $y^2$ is $2y \frac{dy}{dx}$. For the term on the right, we apply the Chain Rule first, followed by the Product Rule:
$$2x + 2y \frac{dy}{dx} = \cos(xy) \cdot \frac{d}{dx}(xy)$$
Applying the Product Rule to the term $xy$ gives:
$$2x + 2y \frac{dy}{dx} = \cos(xy) \cdot \left[ x \frac{dy}{dx} + y \cdot 1 \right]$$
Now, let us expand the right-hand side of the equation:
$$2x + 2y \frac{dy}{dx} = x \cos(xy) \frac{dy}{dx} + y \cos(xy)$$
While solving, let us group all terms containing $\frac{dy}{dx}$ on the left-hand side:
$$2y \frac{dy}{dx} - x \cos(xy) \frac{dy}{dx} = y \cos(xy) - 2x$$
Let us factor out $\frac{dy}{dx}$ on the left-hand side:
$$\frac{dy}{dx} \left[ 2y - x \cos(xy) \right] = y \cos(xy) - 2x$$
Therefore, dividing both sides by the bracketed term gives our final derivative:
$$\frac{dy}{dx} = \frac{y \cos(xy) - 2x}{2y - x \cos(xy)}$$

∴ (y cos(xy) - 2x) / (2y - x cos(xy))

3

If $y = \sin^{-1}x$, prove that $(1-x^2)y_2 - xy_1 = 0$.

Hard

Solution

Let us solve this problem step-by-step. First, let us write down the given function:
$$y = \sin^{-1} x$$
Now, let us find the first derivative $y_1$ with respect to $x$ using our standard formula:
$$y_1 = \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$$
While solving higher-order proofs, a helpful step is to cross-multiply the denominator to avoid using the Quotient Rule for the next derivative:
$$y_1 \sqrt{1-x^2} = 1$$
Now, let us square both sides of the equation to remove the square root:
$$y_1^2 (1-x^2) = 1$$
Let us differentiate both sides of this equation with respect to $x$. Since the left-hand side is a product of two terms, we apply the Product Rule:
$$\frac{d}{dx}\left(y_1^2\right) \cdot (1-x^2) + y_1^2 \cdot \frac{d}{dx}(1-x^2) = \frac{d}{dx}(1)$$$$\left(2y_1 y_2\right)(1-x^2) + y_1^2(-2x) = 0$$
Let us simplify the equation by grouping the terms:
$$2y_1 y_2 (1-x^2) - 2xy_1^2 = 0$$
We can see that $2y_1$ is a common factor in both terms. Since $y_1 \neq 0$, we can safely divide the entire equation by $2y_1$:
$$(1-x^2)y_2 - xy_1 = 0$$
Therefore, the differential relation is proved.

∴ Proved

MCQ Practice

Mixed Level — MCQ

1

The derivative of sec x with respect to x is:

Correct!

Incorrect. Correct: sec x tan x

Solution

Using our standard trigonometric derivatives, the rate of change of $\sec x$ is equal to $\sec x \tan x$.

∴ sec x tan x

2

What is the value of d/dx(ln 5x)?

Correct!

Incorrect. Correct: 1 / x

Solution

Let us use the Chain Rule. First, let us differentiate the outer logarithm function:
$$\frac{d}{d(5x)}(\ln 5x) = \frac{1}{5x}$$
Now, we multiply by the derivative of the inner function:
$$\frac{d}{dx}(5x) = 5$$
Multiplying these together yields:
$$\frac{dy}{dx} = \frac{1}{5x} \cdot 5 = \frac{1}{x}$$

∴ 1 / x

3

The derivative of (tan^-1 x + cot^-1 x) with respect to x is:

Correct!

Incorrect. Correct: 0

Solution

We know from inverse trigonometric properties that:
$$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$$
Since $\frac{\pi}{2}$ is a constant term, its derivative is zero:
$$\frac{d}{dx}\left(\frac{\pi}{2}\right) = 0$$

∴ 0

4

Evaluate the derivative of e^(sin x).

Correct!

Incorrect. Correct: cos x * e^(sin x)

Solution

Let us apply the Chain Rule to find the derivative of this composite function:
$$\frac{d}{dx}\left(e^{\sin x}\right) = e^{\sin x} \cdot \frac{d}{dx}(\sin x)$$
Since the derivative of $\sin x$ is $\cos x$, we get:
$$\frac{dy}{dx} = \cos x \cdot e^{\sin x}$$

∴ cos x * e^(sin x)

5

If y = x^x, then the derivative dy/dx is equal to:

Correct!

Incorrect. Correct: x^x(1 + ln x)

Solution

Let us use Logarithmic Differentiation. First, let us take the natural logarithm of both sides:
$$\ln y = \ln\left(x^x\right) \Rightarrow \ln y = x \ln x$$
Now, we differentiate implicitly with respect to $x$:
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x)$$
Applying the Product Rule to the right-hand side gives:
$$\frac{1}{y} \frac{dy}{dx} = x \cdot \left(\frac{1}{x}\right) + \ln x \cdot (1) = 1 + \ln x$$
Multiplying both sides by $y$ yields:
$$\frac{dy}{dx} = y (1 + \ln x)$$
Finally, substituting $y = x^x$ back into the expression gives:
$$\frac{dy}{dx} = x^x(1 + \ln x)$$

∴ x^x(1 + ln x)

6

What is the derivative of cos(x^3)?

Correct!

Incorrect. Correct: -3x^2 * sin(x^3)

Solution

Let us apply the Chain Rule. We differentiate the outer cosine function first:
$$\frac{d}{d(x^3)}(\cos(x^3)) = -\sin(x^3)$$
Now, we multiply by the derivative of the inner term $x^3$:
$$\frac{d}{dx}(x^3) = 3x^2$$
Combining these gives us:
$$\frac{dy}{dx} = -3x^2 \sin(x^3)$$

∴ -3x^2 * sin(x^3)

7

Find the second derivative (y2) of the function y = x^3 + 5.

Correct!

Incorrect. Correct: 6x

Solution

Let us differentiate successively. First, let us calculate the first derivative:
$$y_1 = \frac{d}{dx}(x^3 + 5) = 3x^2$$
Now, let us differentiate again to find the second derivative:
$$y_2 = \frac{d}{dx}(3x^2) = 6x$$

∴ 6x

8

Differentiate y = √x with respect to x.

Correct!

Incorrect. Correct: 1 / (2√x)

Solution

Let us write $\sqrt{x}$ as $x^{1/2}$ and use the Power Rule:
$$\frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$$

∴ 1 / (2√x)

9

The derivative of a^x with respect to x is:

Correct!

Incorrect. Correct: a^x * ln a

Solution

To find the derivative of $y = a^x$, we can rewrite it using base $e$:
$$y = e^{x \ln a}$$
Now, let us apply the Chain Rule to differentiate with respect to $x$:
$$\frac{dy}{dx} = e^{x \ln a} \cdot \frac{d}{dx}(x \ln a) = e^{x \ln a} \cdot \ln a$$
Substituting $a^x$ back in place of $e^{x \ln a}$ gives:
$$\frac{dy}{dx} = a^x \ln a$$

∴ a^x * ln a

10

Find the derivative of sin^-1(2x) with respect to x.

Correct!

Incorrect. Correct: 2 / √(1 - 4x^2)

Solution

Let us apply the Chain Rule to find this derivative. First, let us write down the derivative of the outer inverse sine function:
$$\frac{d}{du}(\sin^{-1} u) = \frac{1}{\sqrt{1 - u^2}}$$
Substituting $u = 2x$, we get:
$$\frac{dy}{du} = \frac{1}{\sqrt{1 - (2x)^2}} = \frac{1}{\sqrt{1 - 4x^2}}$$
Now, we multiply by the derivative of the inner function $2x$, which is $2$:
$$\frac{dy}{dx} = \frac{2}{\sqrt{1 - 4x^2}}$$

∴ 2 / √(1 - 4x^2)

Short Questions (2 Marks Each)

1

Find $\frac{dy}{dx}$ for the function $y = \tan(4x+5)$.

[2 marks]

Solution

Let us use the Chain Rule. First, let us differentiate the outer tangent function:
$$\frac{d}{d(4x+5)}(\tan(4x+5)) = \sec^2(4x+5)$$
Next, we multiply by the derivative of the inner linear function:
$$\frac{d}{dx}(4x+5) = 4$$
Combining these terms gives us:
$$\frac{dy}{dx} = 4\sec^2(4x+5)$$

∴ 4sec^2(4x+5)

2

Differentiate $y = \ln(\sec x + \tan x)$ with respect to $x$.

[2 marks]

Solution

Let us apply the Chain Rule to this composite logarithmic function:
$$\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$$
We know that the derivatives of these trigonometric terms are:
$$\frac{d}{dx}(\sec x) = \sec x \tan x \quad \text{and} \quad \frac{d}{dx}(\tan x) = \sec^2 x$$
Substituting these in gives:
$$\frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x}$$
Let us factor out $\sec x$ from the numerator:
$$\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x}$$
Canceling the common bracketed term yields:
$$\frac{dy}{dx} = \sec x$$

∴ sec x

3

Find the derivative of $\sqrt{\sin x}$ from first principles.

[2 marks]

Solution

Let us write our function as $f(x) = \sqrt{\sin x}$. Adding an increment $h$ to $x$ gives:
$$f(x+h) = \sqrt{\sin(x+h)}$$
Let us set up the first principles limit:
$$f'(x) = \lim_{h \to 0} \frac{\sqrt{\sin(x+h)} - \sqrt{\sin x}}{h}$$
While solving, let us rationalize the numerator by multiplying with its conjugate:
$$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h(\sqrt{\sin(x+h)} + \sqrt{\sin x})}$$
Now, let us apply the trigonometric C-D formula to the numerator and simplify the limit:
$$f'(x) = \frac{\cos x}{2\sqrt{\sin x}}$$

∴ cos x / (2√sin x)

4

If $y = e^{ax+b}$, find the first derivative $y_1$.

[2 marks]

Solution

Let us use the Chain Rule. First, let us differentiate the outer exponential function, which remains unchanged:
$$\frac{d}{d(ax+b)}\left(e^{ax+b}\right) = e^{ax+b}$$
Next, we multiply by the derivative of the inner term $(ax+b)$:
$$\frac{d}{dx}(ax+b) = a$$
Therefore, multiplying these together gives our first derivative:
$$y_1 = a e^{ax+b}$$

∴ a*e^(ax+b)

5

Differentiate $y = \sin^{-1}(\sqrt{x})$ with respect to $x$.

[2 marks]

Solution

Let us apply the Chain Rule to this composite inverse trigonometric function. First, let us differentiate the outer inverse sine function:
$$\frac{d}{d(\sqrt{x})}(\sin^{-1}(\sqrt{x})) = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{\sqrt{1-x}}$$
Next, we multiply by the derivative of the inner function $\sqrt{x}$:
$$\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$$
Multiplying these together yields:
$$\frac{dy}{dx} = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}}$$

∴ 1 / (2√x(1-x))

6

Find $\frac{dy}{dx}$ if $xy = 1$.

[2 marks]

Solution

Let us differentiate both sides of the equation with respect to $x$ using implicit differentiation.

Applying the Product Rule on the left side gives:
$$x \frac{dy}{dx} + y \cdot 1 = 0$$
Now, let us isolate $\frac{dy}{dx}$ by subtracting $y$ from both sides:
$$x \frac{dy}{dx} = -y$$
Dividing by $x$ gives:
$$\frac{dy}{dx} = -\frac{y}{x}$$

∴ -y/x

7

Find the derivative of $y = \cos^{-1}(2x^2-1)$ with respect to $x$.

[2 marks]

Solution

Let us simplify the function first by using trigonometric substitution. Let us assume that $x = \cos \theta$, which means $\theta = \cos^{-1} x$.

Substituting this into our equation gives:
$$y = \cos^{-1}(2\cos^2 \theta - 1)$$
We know that $2\cos^2 \theta - 1 = \cos 2\theta$. Therefore, we write:
$$y = \cos^{-1}(\cos 2\theta) = 2\theta$$
Let us substitute back $\theta = \cos^{-1} x$:
$$y = 2\cos^{-1} x$$
Now, we differentiate with respect to $x$:
$$\frac{dy}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{2}{\sqrt{1-x^2}}$$

∴ -2 / √(1-x^2)

8

Differentiate $y = \log_2 (x^2)$ with respect to $x$.

[2 marks]

Solution

First, let us simplify the expression by using logarithmic properties. We can bring the power of $2$ to the front:
$$y = 2\log_2 x$$
Next, we apply the change of base formula to convert the logarithm to base $e$:
$$y = 2 \cdot \frac{\ln x}{\ln 2} = \frac{2}{\ln 2} \ln x$$
Now, let us differentiate this expression with respect to $x$:
$$\frac{dy}{dx} = \frac{2}{\ln 2} \cdot \frac{d}{dx}(\ln x) = \frac{2}{\ln 2} \cdot \frac{1}{x} = \frac{2}{x \ln 2}$$

∴ 2 / (x ln 2)

9

Find the third derivative ($y_3$) of $y = x^4$.

[2 marks]

Solution

Let us differentiate successively. First, let us find the first derivative:
$$y_1 = 4x^3$$
Next, let us differentiate again to find the second derivative:
$$y_2 = 4(3x^2) = 12x^2$$
Finally, let us differentiate one more time to find the third derivative:
$$y_3 = 12(2x) = 24x$$

∴ 24x

10

Find the derivative of $\sin x$ with respect to $\cos x$.

[2 marks]

Solution

Let us assume that $u = \sin x$ and $v = \cos x$. We wish to find $\frac{du}{dv}$.

First, let us differentiate both terms with respect to $x$:
$$\frac{du}{dx} = \cos x \quad \text{and} \quad \frac{dv}{dx} = -\sin x$$
Now, we can find the derivative by taking the ratio of these values:
$$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\cos x}{-\sin x} = -\cot x$$

∴ -cot x

Long Questions (Board Style)

1

Derive the derivative of $\tan x$ from First Principles.

[3 marks]

Solution

Let us write down our function as $f(x) = \tan x$.

Step 1 — Introduce the increment:
Let us give a small increment $h$ to $x$, which gives us:
$$f(x+h) = \tan(x+h)$$

Step 2 — Set up the limit:
We know that the first principles definition is:
$$f'(x) = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h}$$

Step 3 — Convert to sine and cosine:
Let us rewrite the tangent functions in terms of sine and cosine:
$$f'(x) = \lim_{h \to 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}}{h}$$
Now, let us find a common denominator for the numerator:
$$f'(x) = \lim_{h \to 0} \frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{h \cos(x+h)\cos x}$$

Step 4 — Apply trigonometric identities:
We know the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Let us apply this with $A = x+h$ and $B = x$:
$$f'(x) = \lim_{h \to 0} \frac{\sin(x+h-x)}{h \cos(x+h)\cos x} = \lim_{h \to 0} \frac{\sin h}{h \cos(x+h)\cos x}$$

Step 5 — Separate and evaluate the limit:
Let us split the limit into two parts:
$$f'(x) = \left( \lim_{h \to 0} \frac{\sin h}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{\cos(x+h)\cos x} \right)$$
Using the standard limit $\lim_{h \to 0} \frac{\sin h}{h} = 1$, we get:
$$f'(x) = 1 \cdot \frac{1}{\cos(x+0)\cos x} = \frac{1}{\cos^2 x} = \sec^2 x$$

∴ sec^2 x

2

If $y = (\sin x)^{\ln x}$, find $\frac{dy}{dx}$ using logarithmic differentiation.

[3 marks]

Solution

Since the variable appears in both the base and the exponent, let us apply Logarithmic Differentiation.

Step 1 — Take the natural logarithm:
Taking $\ln$ on both sides of the equation gives:
$$\ln y = \ln\left((\sin x)^{\ln x}\right)$$
We can bring the exponent to the front:
$$\ln y = \ln x \cdot \ln(\sin x)$$

Step 2 — Differentiate implicitly:
Differentiating both sides with respect to $x$ using the Product Rule on the right-hand side gives:
$$\frac{1}{y} \frac{dy}{dx} = \ln x \cdot \frac{d}{dx}(\ln(\sin x)) + \ln(\sin x) \cdot \frac{d}{dx}(\ln x)$$
Let us compute the derivatives on the right-hand side:
$$\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$$
$$\frac{d}{dx}(\ln x) = \frac{1}{x}$$
Substituting these values back in gives:
$$\frac{1}{y} \frac{dy}{dx} = \ln x \cdot \cot x + \frac{\ln(\sin x)}{x}$$

Step 3 — Isolate dy/dx:
Multiplying both sides by $y$ gives:
$$\frac{dy}{dx} = y \cdot \left[ \ln x \cot x + \frac{\ln(\sin x)}{x} \right]$$
Now, let us substitute the original expression $y = (\sin x)^{\ln x}$ back into the equation:
$$\frac{dy}{dx} = (\sin x)^{\ln x} \cdot \left[ \ln x \cot x + \frac{\ln(\sin x)}{x} \right]$$

∴ (sin x)^(ln x) * [ln x cot x + ln(sin x)/x]

3

If $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$, find $\frac{dy}{dx}$.

[3 marks]

Solution

Since $x$ and $y$ are both expressed in terms of the parameter $\theta$, let us perform Parametric Differentiation.

Step 1 — Find dx/dθ:
$$\frac{dx}{d\theta} = a(1 + \cos \theta)$$

Step 2 — Find dy/dθ:
$$\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a\sin \theta$$

Step 3 — Calculate dy/dx:
We know that the parametric derivative is given by:
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$
Substituting our derivatives into the formula gives:
$$\frac{dy}{dx} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$$
Let us simplify this using half-angle trigonometric formulas:
$$\sin \theta = 2\sin(\theta/2)\cos(\theta/2) \quad \text{and} \quad 1 + \cos \theta = 2\cos^2(\theta/2)$$
Substituting these into the fraction yields:
$$\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2)$$

∴ tan(θ/2)

4

Prove that the derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$.

[3 marks]

Solution

Let us prove this formula. First, let us assume that:
$$y = \sin^{-1} x$$
This can be rewritten in terms of the sine function as:
$$\sin y = x$$
Now, let us differentiate both sides with respect to $x$ using implicit differentiation:
$$\cos y \cdot \frac{dy}{dx} = 1$$
Let us isolate $\frac{dy}{dx}$ by dividing both sides by $\cos y$:
$$\frac{dy}{dx} = \frac{1}{\cos y}$$
Using the trigonometric identity $\cos^2 y + \sin^2 y = 1$, we can express $\cos y$ as:
$$\cos y = \sqrt{1 - \sin^2 y}$$
Substituting $\sin y = x$ into this equation gives:
$$\cos y = \sqrt{1 - x^2}$$
Therefore, substituting this expression back into our derivative formula yields:
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$

∴ 1 / √(1 - x^2)

5

Find $\frac{dy}{dx}$ if $x^y \cdot y^x = k$, where $k$ is a constant.

[3 marks]

Solution

Since the exponents contain variables, let us simplify the equation by taking the natural logarithm of both sides:
$$\ln\left(x^y \cdot y^x\right) = \ln k$$
Using logarithmic properties, we can expand the left side as:
$$y \ln x + x \ln y = \ln k$$
Now, let us differentiate both sides with respect to $x$. Since both terms on the left are products, we apply the Product Rule to each:
$$\left[ y \cdot \left(\frac{1}{x}\right) + \ln x \cdot \frac{dy}{dx} \right] + \left[ x \cdot \left(\frac{1}{y} \frac{dy}{dx}\right) + \ln y \cdot 1 \right] = 0$$
Let us group the terms containing $\frac{dy}{dx}$ together:
$$\left( \ln x + \frac{x}{y} \right) \frac{dy}{dx} = -\left( \frac{y}{x} + \ln y \right)$$$$\left( \frac{y \ln x + x}{y} \right) rac{dy}{dx} = -\left( \frac{y + x \ln y}{x} \right)$$
Isolating $\frac{dy}{dx}$ gives our final derivative:
$$\frac{dy}{dx} = -\frac{y(y + x \ln y)}{x(x + y \ln x)}$$

∴ -y(y + x ln y) / [x(x + y ln x)]

6

Derive the derivative of $\ln x$ from First Principles.

[3 marks]

Solution

Let us assume that our function is $f(x) = \ln x$.

Step 1 — Introduce the increment:
Let us give a small increment $h$ to $x$:
$$f(x+h) = \ln(x+h)$$

Step 2 — Set up the limit:
We know that the first principles formula is:
$$f'(x) = \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h}$$

Step 3 — Apply logarithmic properties:
Using the property $\ln A - \ln B = \ln(A/B)$, we simplify the numerator:
$$f'(x) = \lim_{h \to 0} \frac{\ln\left(\frac{x+h}{x}\right)}{h} = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h}$$

Step 4 — Use substitution to evaluate the limit:
Let us substitute $u = \frac{h}{x}$, which implies $h = xu$. As $h \to 0$, the variable $u$ also approaches $0$. Substituting these into the limit gives:
$$f'(x) = \lim_{u \to 0} \frac{\ln(1 + u)}{xu} = \frac{1}{x} \cdot \lim_{u \to 0} \frac{\ln(1 + u)}{u}$$
We know that the standard limit is $\lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1$. Therefore, we find:
$$f'(x) = \frac{1}{x} \cdot 1 = \frac{1}{x}$$

∴ 1/x

7

Solve the infinite radical equation $y = \sqrt{\tan x + \sqrt{\tan x + \dots}}$ to find $\frac{dy}{dx}$.

[3 marks]

Solution

Since the pattern of square roots repeats infinitely, we can substitute the nested terms with $y$:
$$y = \sqrt{\tan x + y}$$
To make the equation easier to differentiate, let us square both sides:
$$y^2 = \tan x + y$$
Now, let us differentiate both sides with respect to $x$ using implicit differentiation:
$$2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$$
Let us group all terms containing $\frac{dy}{dx}$ on the left-hand side:
$$2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x$$
Factoring out $\frac{dy}{dx}$ gives:
$$\frac{dy}{dx} (2y - 1) = \sec^2 x$$
Dividing both sides by $(2y-1)$ yields our final derivative:
$$\frac{dy}{dx} = \frac{\sec^2 x}{2y - 1}$$

∴ sec^2 x / (2y - 1)

8

If $y = A e^{nx} + B e^{-nx}$, prove that $y_2 - n^2 y = 0$.

[3 marks]

Solution

Let us differentiate the given function successively. First, let us write down the equation:
$$y = A e^{nx} + B e^{-nx}$$
Now, let us differentiate with respect to $x$ using the Chain Rule to find the first derivative $y_1$:
$$y_1 = \frac{dy}{dx} = A \cdot n e^{nx} + B \cdot (-n) e^{-nx} = A n e^{nx} - B n e^{-nx}$$
Let us differentiate a second time to find the second derivative $y_2$:
$$y_2 = \frac{d^2y}{dx^2} = A n \cdot n e^{nx} - B n \cdot (-n) e^{-nx}$$$$y_2 = A n^2 e^{nx} + B n^2 e^{-nx}$$
We can factor out $n^2$ from both terms on the right-hand side:
$$y_2 = n^2 \left( A e^{nx} + B e^{-nx} \right)$$
Notice that the term inside the parentheses is equal to our original function $y$. Substituting $y$ back in gives:
$$y_2 = n^2 y$$
Subtracting $n^2 y$ from both sides completes our proof:
$$y_2 - n^2 y = 0$$

∴ Proved

9

Differentiate $y = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ with respect to $x$.

[3 marks]

Solution

Let us simplify the algebraic expression inside the parentheses first using half-angle trigonometric formulas. We know that:
$$\sin x = 2\sin(x/2)\cos(x/2) \quad \text{and} \quad 1+\cos x = 2\cos^2(x/2)$$
Substituting these formulas into our expression gives:
$$\frac{\sin x}{1+\cos x} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2)$$
Now, let us substitute this simplified term back into our function:
$$y = \tan^{-1}(\tan(x/2))$$
Using inverse properties, we can simplify this to:
$$y = \frac{x}{2}$$
Differentiating this linear expression with respect to $x$ yields our final result:
$$\frac{dy}{dx} = \frac{1}{2}$$

∴ 1/2

10

Differentiate the function $\sin(x^2)$ with respect to $x^2$.

[3 marks]

Solution

Let us introduce a new variable to simplify the problem. Let us assume that $u = x^2$ is our independent variable.

Now, we can rewrite our function in terms of $u$:
$$y = \sin u$$
The question asks us to differentiate $\sin(x^2)$ with respect to $x^2$, which means we need to calculate $\frac{dy}{du}$.

Let us differentiate $y$ directly with respect to $u$:
$$\frac{dy}{du} = \cos u$$
Finally, let us substitute $u = x^2$ back into our derivative:
$$\frac{dy}{du} = \cos(x^2)$$

∴ cos(x^2)

Past NEB Board Questions

1

Find the derivative of $\cos 3x$ from first principles.

NEB 2079 [3 marks]

Solution

Let us write our function as $f(x) = \cos 3x$.

Step 1 — Introduce the increment:
Let us give a small increment $h$ to $x$:
$$f(x+h) = \cos(3(x+h)) = \cos(3x+3h)$$

Step 2 — Set up the limit:
We know that the first principles definition is:
$$f'(x) = \lim_{h \to 0} \frac{\cos(3x+3h) - \cos 3x}{h}$$

Step 3 — Apply the trigonometric C-D formula:
Using the identity $\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$ with $C = 3x+3h$ and $D = 3x$:
$$f'(x) = \lim_{h \to 0} \frac{-2\sin\left(\frac{3x+3h+3x}{2}\right)\sin\left(\frac{3x+3h-3x}{2}\right)}{h}$$$$f'(x) = \lim_{h \to 0} \frac{-2\sin\left(3x+\frac{3h}{2}\right)\sin\left(\frac{3h}{2}\right)}{h}$$

Step 4 — Separate and evaluate the limit:
To match our standard limit, let us divide and multiply the denominator by $3/2$:
$$f'(x) = \lim_{h \to 0} -2\sin\left(3x+\frac{3h}{2}\right) \cdot \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2} \cdot \frac{2}{3}}$$$$f'(x) = -3 \cdot \lim_{h \to 0} \sin\left(3x+\frac{3h}{2}\right) \cdot \lim_{h \to 0} \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}}$$
Since the second limit is equal to $1$, we substitute $h=0$ into the first term:
$$f'(x) = -3 \sin(3x) \cdot 1 = -3\sin 3x$$

∴ -3sin 3x

2

Find $\frac{dy}{dx}$ if $y = x^{\sin x}$ using logarithmic differentiation.

NEB 2078 [3 marks]

Solution

Since the variable appears in both the base and the exponent, let us apply Logarithmic Differentiation.

Step 1 — Take the natural logarithm:
Taking $\ln$ on both sides of our equation gives:
$$\ln y = \ln\left(x^{\sin x}\right) = \sin x \ln x$$

Step 2 — Differentiate implicitly:
Let us differentiate both sides with respect to $x$ using the Product Rule on the right side:
$$\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(\sin x)$$$$$\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \left(\frac{1}{x}\right) + \ln x \cdot \cos x$$

Step 3 — Solve for dy/dx:
Multiplying both sides by $y$ gives:
$$\frac{dy}{dx} = y \cdot \left[ \frac{\sin x}{x} + \cos x \ln x \right]$$
Now, let us substitute our original expression $y = x^{\sin x}$ back into the equation:
$$\frac{dy}{dx} = x^{\sin x} \cdot \left[ \frac{\sin x}{x} + \cos x \ln x \right]$$

∴ x^(sin x) * [sin(x)/x + cos(x) ln x]

3

If $y = \tan^{-1}x$, prove that $(1+x^2)y_2 + 2xy_1 = 0$.

NEB 2077 [3 marks]

Solution

Let us differentiate successively. First, let us write down our function:
$$y = \tan^{-1} x$$
Now, let us find the first derivative $y_1$ using our standard formula:
$$y_1 = \frac{1}{1+x^2}$$
While solving, let us cross-multiply to avoid having to use the Quotient Rule for the next derivative:
$$y_1 (1+x^2) = 1$$
Let us differentiate both sides of this equation with respect to $x$ using the Product Rule on the left side:
$$\frac{d}{dx}(y_1) \cdot (1+x^2) + y_1 \cdot \frac{d}{dx}(1+x^2) = \frac{d}{dx}(1)$$$$y_2 (1+x^2) + y_1(2x) = 0$$
Arranging the terms completes our proof:
$$(1+x^2)y_2 + 2xy_1 = 0$$

∴ Proved

Chapter Test — Full Mixed Paper

Complete this examination-style test paper in 45 minutes under quiet exam conditions.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

What is the derivative of csc x with respect to x?

Correct!

Incorrect. Correct: -csc x * cot x

Solution

Using our standard trigonometric derivative properties, we find:
$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

∴ -csc x * cot x

2

The derivative of a constant k is:

Correct!

Incorrect. Correct: 0

Solution

Since a constant has no variables, its rate of change is zero:
$$\frac{d}{dx}(k) = 0$$

∴ 0

3

The value of the derivative of e^x at x = 0 is:

Correct!

Incorrect. Correct: 1

Solution

First, let us write down the derivative of $e^x$, which is $e^x$. Substituting $x = 0$ gives:
$$\left. \frac{d}{dx}\left(e^x\right) \right|_{x=0} = e^0 = 1$$

∴ 1

2 Section B — Short Questions (2 Marks Each)

1

Find $\frac{dy}{dx}$ if $y = \ln(\sin 2x)$.

[2 marks]

Solution

Let us apply the Chain Rule. First, let us differentiate the outer logarithm function:
$$\frac{d}{d(\sin 2x)}(\ln(\sin 2x)) = \frac{1}{\sin 2x}$$
Now, we multiply by the derivative of the inner term $\sin 2x$:
$$\frac{d}{dx}(\sin 2x) = 2\cos 2x$$
Multiplying these together yields:
$$\frac{dy}{dx} = \frac{2\cos 2x}{\sin 2x} = 2\cot 2x$$

∴ 2cot 2x

2

Differentiate $x^2 + y^2 = 25$ implicitly.

[2 marks]

Solution

Let us differentiate both sides with respect to $x$:
$$2x + 2y \frac{dy}{dx} = 0$$
Let us isolate the term containing $\frac{dy}{dx}$:
$$2y \frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$$

∴ -x/y

3 Section C — Long Questions (3 Marks Each)

1

If $y = x^{\sin x}$, find $\frac{dy}{dx}$.

[3 marks]

Solution

Since the variable is in both the base and the exponent, let us take the natural logarithm of both sides:
$$\ln y = \sin x \ln x$$
Now, we differentiate implicitly using the Product Rule on the right side:
$$\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \left(\frac{1}{x}\right) + \ln x \cdot \cos x$$
Multiplying both sides by $y$ and substituting $y = x^{\sin x}$ back in yields our final derivative:
$$\frac{dy}{dx} = x^{\sin x} \cdot \left[ \frac{\sin x}{x} + \cos x \ln x \right]$$

∴ x^(sin x) * [sin(x)/x + cos(x) ln x]

One-Page Revision Cheat Sheet

1

First Principles Limits: When evaluating trigonometric functions from first principles, always aim to construct the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

2

Trigonometric signs: Remember that any trigonometric function starting with 'co' (cosine, cotangent, cosecant) has a negative sign in its derivative.

3

Logarithmic property: For variable-exponent expressions, take the natural log of both sides first to simplify the exponent structure.

4

Chain Rule ordering: Differentiate nested functions from the outside in, multiplying the individual derivatives together.

5

Implicit grouping: Differentiate term-by-term and group all terms containing $\frac{dy}{dx}$ on one side to solve for the derivative.

6

Successive Differentiation: To prove differential equations, calculate the first derivative, simplify it by cross-multiplying to avoid quotient fractions, and differentiate again.