CBSE Board Examination 2025 · Section B · 2 marks

Q21: Differentiate $2^{\cos^2 x}$ w.r.t. $\cos^2 x$ ⁠, or $\frac{dy}{dx}$ when $\tan^{-1}(x^2 + y^2) = a^2$

Board 2025Ch 5 · Continuity and Differentiability

The question

(a) Differentiate $2^{\cos^2 x}$ w.r.t $\cos^2 x$ ⁠.

(b) If $\tan^{-1}(x^2 + y^2) = a^2$ ⁠, then find $\dfrac{dy}{dx}$ ⁠.

Exam answer

(a)

Let $u = 2^{\cos^2 x}$ and $v = \cos^2 x$ ⁠.

Why?

We want the derivative of one function with respect to another function. Both $u$ and $v$ are functions of $x$ ⁠, so we differentiate each of them with respect to $x$ and then divide.

\begin{aligned} \frac{du}{dx} &= 2^{\cos^2 x}\log 2 \\ &\quad\times(-2\cos x\sin x) \end{aligned}

Why?

The variable is in the exponent, so use the exponential rule with the chain rule ( $k$ a constant):

\frac{d}{dx}k^{t} = k^{t}\log k\cdot\frac{dt}{dx}

Here $k = 2$ and $t = \cos^2 x$ ⁠. By the chain rule again,

\begin{aligned} \frac{d}{dx}\cos^2 x &= 2\cos x\,(-\sin x) \\ &= -2\cos x\sin x \end{aligned}

(Outside: $t^2$ gives $2t$ ⁠, with $t = \cos x$ ⁠. Inside: $\cos x$ gives $-\sin x$ ⁠.) This is also $\frac{dv}{dx}$ ⁠, the next line.

Watch out: $k^{t}$ is not a power of $t$ ⁠. Do not write $t\,k^{t - 1}$ ⁠.

\frac{dv}{dx} = -2\cos x\sin x

\frac{du}{dv} = \frac{du/dx}{dv/dx}

Why?

This is the chain rule read backwards:

\frac{du}{dx} = \frac{du}{dv}\cdot\frac{dv}{dx}

Divide both sides by $\frac{dv}{dx}$ ⁠, which is allowed where $\cos x\sin x \ne 0$ ⁠.

Cancelling $-2\cos x\sin x$ ⁠,

Why?

Put both derivatives into the fraction. The same factor sits on the top and the bottom:

\begin{aligned} \frac{du}{dv} &= \frac{2^{\cos^2 x}\log 2\,(-2\cos x\sin x)}{-2\cos x\sin x} \\ &= 2^{\cos^2 x}\log 2 \end{aligned}

$\therefore\ \frac{du}{dv} = 2^{\cos^2 x}\log 2$ ⁠.

(b)

\begin{aligned} &\tan^{-1}(x^2 + y^2) = a^2 \\ \Rightarrow\ &x^2 + y^2 = \tan a^2 \end{aligned}

Why?

Take $\tan$ of both sides. Since $\tan(\tan^{-1}t) = t$ for every real $t$ ⁠, the inverse function disappears.

$a$ is a constant, so $\tan a^2$ is just a number. This makes the differentiation much shorter.

Differentiating both sides w.r.t. $x$ ( $\tan a^2$ is a constant),

2x + 2y\frac{dy}{dx} = 0

Why?

$y$ is a function of $x$ ⁠, so differentiate $y^2$ by the chain rule:

\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}

The derivative of a constant is $0$ ⁠.

Watch out: writing $\frac{d}{dx}(y^2) = 2y$ loses the $\frac{dy}{dx}$ ⁠.

2y\frac{dy}{dx} = -2x

Why?

Move $2x$ to the right side, then divide both sides by $2y$ (where $y \ne 0$ ⁠).

\begin{aligned} 2x + 2y\frac{dy}{dx} &= 0 \\ 2y\frac{dy}{dx} &= -2x \\ \frac{dy}{dx} &= \frac{-2x}{2y} \\ &= -\frac{x}{y} \quad (\text{cancel } 2) \end{aligned}

$\therefore\ \dfrac{dy}{dx} = -\dfrac{x}{y}$ ⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative w.r.t. a function; implicit functions.

Question type: (a) Parametric form; (b) Implicit

How to spot it: click to see

(a) Type 8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Here: Both $u = 2^{\cos^2 x}$ and $v = \cos^2 x$ are given through $x$ ⁠, so $x$ plays the parameter: find $\frac{du}{dx}$ and $\frac{dv}{dx}$ ⁠, and divide. The factor $-2\cos x\sin x$ cancels.

(b) Type 5. Implicit: $x$ and $y$ mixed in one equation

Spot it: an equation such as $x^2 + xy + y^2 = 100$ that you cannot easily solve for $y$ ⁠.

Here: $x$ and $y$ are mixed inside $\tan^{-1}$ ⁠. Take $\tan$ of both sides, $x^2 + y^2 = \tan a^2$ ⁠, a constant; differentiate each term, $y^2$ giving $2y\frac{dy}{dx}$ ⁠.

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Continuity and Differentiability in past papers: question 5 of 35 · see them all

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Q21: Differentiate $2^{\cos^2 x}$ w.r.t. $\cos^2 x$ ⁠, or $\frac{dy}{dx}$ when $\tan^{-1}(x^2 + y^2) = a^2$