(a) Differentiate $2^{\cos^2 x}$ with respect to $\cos^2 x$. OR (b) If $\tan^{-1}(x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.

(a) Differentiate 2cos^2 x with respect to cos² x. OR (b) If tan⁻¹(x² + y²) = a², then find (dy)/(dx).

Question

2^{\cos^2 x}

(OR)

(b) If

\tan^{-1}(x^2 + y^2) = a^2

, then find

\frac{dy}{dx}

CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Derivative Evaluation

Derivative Evaluation

To evaluate a derivative means to find $f'(a)$ — a single number that tells you how fast $f$ is changing right at $x=a$ . Think of a speedometer: it doesn't report how far you've travelled, only how fast your position is changing at this instant. That instantaneous rate of change is exactly what $f'(a)$ measures.

The geometric picture

On the curve $y=f(x)$ , pick a point $P$ and a nearby point $Q$ . The straight line through them — the secant — has slope equal to the average rate of change between $P$ and $Q$ . Now slide $Q$ toward $P$ : the secant rotates into the tangent line that just touches the curve at $P$ , and its slope is $f'(a)$ .

Note

$f'(a)$ is the slope of the tangent to $y=f(x)$ at $x=a$ — how steep the curve is right there.

The limit definition

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Here $h$ is a tiny step from $a$ to $a+h$ , the numerator is the matching change in height, and the ratio is a secant slope. As $h\to 0$ the secant becomes the tangent. An equivalent form is

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

When this limit exists, $f$ is differentiable at $a$ (which forces continuity there).

Watch out

Continuity alone is not enough. $f(x)=|x|$ is continuous at $0$ , but its left slope $-1$ and right slope $+1$ disagree, so $f'(0)$ does not exist — a corner has no single tangent.

A worked evaluation

For $f(x)=x^2$ at $x=3$ :

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} (6 + h) = 6.

So the tangent at $x=3$ has slope $6$ .

From a number to a function …

Part (b)Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

You can differentiate $y=x^2+3x$ term by term because $y$ is written explicitly in terms of $x$ . But an equation like $x^2+y^2=25$ , or $x^3+y^3=6xy$ , does not give $y$ by itself — solving for $y$ is messy or downright impossible.

Implicit differentiation finds $\dfrac{dy}{dx}$ without isolating $y$ : treat $y$ as an unknown function of $x$ , differentiate the whole equation as it stands, then solve for $\dfrac{dy}{dx}$ .

The one key move: $y$ is really $y(x)$

Wherever $y$ appears, picture $y(x)$ hiding inside. Differentiating a $y$ -term therefore needs the chain rule, which tacks on a factor of $\dfrac{dy}{dx}$ :

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

That extra $\dfrac{dy}{dx}$ on every $y$ -term is the whole trick.

The procedure

x

Worked example

For $x^2+y^2=25$ :

2x+2y\frac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=-\frac{x}{y}.

The answer naturally contains both $x$ and $y$ — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for $y$ first. …

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Part (a): $\dfrac{d}{d(\cos^2 x)}2^{\cos^2 x}=2^{\cos^2 x}\ln 2$ . Part (b): $\dfrac{dy}{dx}=-\dfrac{x}{y}$ .

Part (a)

Idea. "Differentiate with respect to $\cos^2 x$ " means treat $\cos^2 x$ as the variable — substitute $u=\cos^2 x$ .

Let $u=\cos^2 x$ ; the expression becomes $2^u.$
Standard rule: $\dfrac{d}{du}a^u=a^u\ln a.$ With $a=2$ :

\frac{d}{du}2^u=2^u\ln 2.

Substitute back $u=\cos^2 x$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 130 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$e^{-x} + e^{-y} = 2$
›Reveal solutionSolution
To find $\frac{dy}{dx}$ for an implicitly defined function, we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule. The result is $\boxed{-e^{y-x}}$ .

When an equation relates $x$ and $y$ but does not explicitly express $y$ as a function of $x$ (like $y = f(x)$ ), we use a technique called implicit differentiation to find $\frac{dy}{dx}$ . The core idea is that even though $y$ isn't isolated, it is still a function of $x$ .

This means that when we differentiate a term involving $y$ with respect to $x$ , we must apply the chain rule. For example, if we differentiate $g(y)$ with respect to $x$ , we get $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ . This $\frac{dy}{dx}$ term is crucial and often the source of errors if overlooked.

Let's apply this to the given equation.

Differentiate both sides of the equation with respect to $x$ . The given equation is $e^{-x} + e^{-y} = 2$ . We apply the derivative operator $\frac{d}{dx}$ to every term:

$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$

Evaluate each derivative.

For the first term, $\frac{d}{dx}(e^{-x})$ :

Using the chain rule, if $u = -x$ , then $\frac{du}{dx} = -1$ .

So, $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$ .

For the second term, $\frac{d}{dx}(e^{-y})$ :

This is where implicit differentiation comes in. We treat $y$ as a function of $x$ .

Using the chain rule, if $v = -y$ , then $\frac{dv}{dx} = \frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$ .

So, $\frac{d}{dx}(e^{-y}) = e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot \left(-\frac{dy}{dx}\right) = -e^{-y} \frac{dy}{dx}$ .

Watch out

A common mistake is to forget the $\frac{dy}{dx}$ term when differentiating expressions involving $y$ with respect to $x$ . Remember, $y$ is a function of $x$ .

For the right-hand side, $\frac{d}{dx}(2)$ : …
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CBSE 2026Set 65/3/11 markMCQ
$\sin^{-1} x = y$
›Reveal solutionSolution
The derivative of $\sin^{-1} x$ is found by implicit differentiation of $x = \sin y$ , giving $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$ , which matches option (B) $\cos y$ only if we interpret it as $\sec y$ — but careful: the correct form is $\frac{1}{\sqrt{1-x^2}}$ , and among the given choices, (B) $\cos y$ is actually $\frac{1}{\cos y}$ ? No — let's check: $\cos y = \sqrt{1-x^2}$ , so $\frac{1}{\cos y} = \sec y$ , which is option (D). The final answer is (D) $\sec y$ .

The core idea: when you have an inverse trigonometric function, the easiest way to differentiate it is to rewrite it as a direct trigonometric equation and then use implicit differentiation. This avoids memorising a dozen formulas and builds from what you already know — the derivative of $\sin$ and the chain rule.

Let $\sin^{-1} x = y$ . This means $x = \sin y$ , and importantly, $y$ is restricted to $[-\pi/2, \pi/2]$ so that $\cos y \ge 0$ .

Start with the relation:

$x = \sin y$

Differentiate both sides with respect to $x$ . Remember $y$ is a function of $x$ , so we use the chain rule on the right:

$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$ $1 = \cos y \cdot \frac{dy}{dx}$

Solve for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{1}{\cos y}$

Now, $\cos y$ can be expressed in terms of $x$ . Since $\sin y = x$ , we use the identity $\sin^2 y + \cos^2 y = 1$ :

$\cos^2 y = 1 - \sin^2 y = 1 - x^2$ $\cos y = \sqrt{1 - x^2} \quad (\text{positive because } y \in [-\pi/2, \pi/2])$

Therefore: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ …
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CBSE 2026Set V11 markMCQ
$x - y = \pi$
(a) $\pi$
(b) $-\pi$
(c) $1$
(d) $-1$
›Reveal solutionSolution

Differentiating the constant-difference relation gives $\dfrac{dy}{dx}=1$ ; answer (c).

Differentiate $x-y=\pi$ (a constant) with respect to $x$ : …
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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\log x^n) =$
(a) $\frac{1}{x^n}$
(b) $n$
(c) $\frac{1}{x}$
(d) $\frac{n}{x}$
›Reveal solutionSolution

$\frac{d}{dx}(\log x^n) = \frac{n}{x}$ .

First simplify with the log power rule:

$\log x^n = n\log x.$

Differentiate: …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(e^{x-a}) =$
(a) $e^{x-a}$
(b) $(x-a)e^{x-a}$
(c) $e^x$
(d) $-e^{x-a}$
›Reveal solutionSolution

$\frac{d}{dx}\left(e^{x-a}\right) = e^{x-a}$ .

Here $a$ is a constant. Let $u = x-a$ , so $\frac{du}{dx} = 1$ .

By the chain rule, …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x}) =$
(a) $\frac{\pi}{2}$
(b) $0$
(c) $1$
(d) $\pi$
›Reveal solutionSolution

The derivative is $0$ because the sum is a constant.

For any real argument $t$ , $\tan^{-1}t + \cot^{-1}t = \frac{\pi}{2}$ . Taking $t=\sqrt{x}$ ,

$\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x} = \frac{\pi}{2}.$

…

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(2\tan^{-1}x) =$
(a) $\frac{1}{1+x^2}$
(b) $\frac{2}{1+x^2}$
(c) $\frac{1}{2(1+x^2)}$
(d) $\frac{1}{1-x^2}$
›Reveal solutionSolution

$\frac{d}{dx}(2\tan^{-1}x) = \frac{2}{1+x^2}$ .

Using the standard derivative $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ and the constant multiple rule: …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\left\{\lim_{x\to 0}\frac{x^5-a^5}{x-a}\right\} =$
(a) $a$
(b) $0$
(c) $5a^4$
(d) $5$
›Reveal solutionSolution

The inner limit is a constant, so the derivative is $0$ .

Evaluate the limit first (it does not depend on $x$ after the limit is taken):

$\lim_{x\to a}\frac{x^5-a^5}{x-a} = 5a^4$ …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\cot^{-1}x) =$
(a) $\frac{1}{1+x^2}$
(b) $\frac{-1}{1+x^2}$
(c) $\frac{1}{x}$
(d) $\frac{-1}{x}$
›Reveal solutionSolution

$\frac{d}{dx}\cot^{-1}x = \frac{-1}{1+x^2}$ .

This is a standard result. From $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ , differentiating gives …

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CBSE 2026Set A1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$
(a) $\frac{1}{2y-1}$
(b) $\frac{\cos x}{2y-1}$
(c) $\frac{\sin x}{2y-1}$
(d) $\frac{2y-1}{\cos x}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y-1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ , so

$y^2 = \sin x + y.$

Differentiate both sides implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$

Collect $\frac{dy}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$x^n + y^n = a^n$
(a) $-\frac{x^{n-1}}{y^{n-1}}$
(b) $\frac{x^{n-1}}{y^{n-1}}$
(c) $-\frac{y^{n-1}}{x^{n-1}}$
(d) $nx^{n-1}$
›Reveal solutionSolution

$\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}$ .

Differentiate $x^n + y^n = a^n$ implicitly ( $a$ constant):

$n x^{n-1} + n y^{n-1}\frac{dy}{dx} = 0.$

Solve: …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sin^{-1}(3x - 4x^3) =$
(a) $\frac{3}{\sqrt{1-x^2}}$
(b) $\frac{-3}{\sqrt{1-x^2}}$
(c) $\frac{1}{\sqrt{1-x^2}}$
(d) $\frac{-1}{\sqrt{1-x^2}}$
›Reveal solutionSolution

$\frac{d}{dx}\sin^{-1}(3x-4x^3) = \frac{3}{\sqrt{1-x^2}}$ .

Use the identity (for the principal branch, $-\frac{1}{2}\le x\le\frac{1}{2}$ ):

$\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x,$

which follows from $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ with $x=\sin\theta$ .

Differentiate: …

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Q.(a) Differentiate 2cos⁡2x2^{\cos^2 x}2cos2x with respect to cos⁡2x\cos^2 xcos2x.

Part (a)

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Q.(a) Differentiate $2^{\cos^2 x}$ with respect to $\cos^2 x$ .

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