The derivative of $2^x$ w.r.t. $3^x$ is: (A) $\left(\dfrac{3}{2}\right)^x \dfrac{\log 2}{\log 3}$ (B) $\left(\dfrac{2}{3}\right)^x \dfrac{\log 3}{\log 2}$ (C) $\left(\dfrac{2}{3}\right)^x \dfrac{\log 2}{\log 3}$ (D) $\left(\dfrac{3}{2}\right)^x \dfrac{\log 3}{\log 2}$

> [!TLDR] > To find the derivative of $f(x)$ with respect to $g(x)$, we compute the ratio of their individual derivatives with respect to $x$. For $f(x) = 2^x$ and $g(x) = 3^x$, this yields $\left(\frac{2}{3}\right)^x \frac{\log 2}{\log 3}$. When asked to find the derivative of one function with respect to another, say the derivative of $u = f(x)$ with respect to $v = g(x)$, we are essentially looking for $\frac{du}{dv}$. Since both $u$ and $v$ are functions of a common variable $x$, we can use the chain rule. The chain rule allows us to express $\frac{du}{dv}$ in terms of derivatives with respect to $x$: $\frac{du}{dv} = \frac{du/dx}{dv/dx}$ This means we need to: 1. Find the derivative of the first function ($2^x$) with respect to $x$. 2. Find the derivative of the second function ($3^x$) with respect to $x$. 3. Divide the first result by the second result. Let's proceed with the steps. 1. **Identify the functions**: Let $u = 2^x$ be the function whose derivative is required. Let $v = 3^x$ be the function with respect to which we are differentiating. We need to find $\frac{du}{dv}$. 2. **Differentiate $u$ with respect to $x$**: We use the standard differentiation formula for exponential functions. > [!FORMULA] > The derivative of $a^x$ with respect to $x$ is $\frac{d}{dx}(a^x) = a^x \log a$. (Here, $\log a$ denotes the natural logarithm, $\ln a$.) Applying this to $u = 2^x$: $\frac{du}{dx} = \frac{d}{dx}(2^x) = 2^x \log 2$. 3. **Differentiate $v$ with respect to $x$**: Similarly, applying the formula to $v = 3^x$: $\frac{dv}{dx} = \frac{d}{dx}(3^x) = 3^x \log 3$. …

The derivative of 2x w.r.t. 3x is: (A) ((3)/(2))x (log 2)/(log 3) (B) ((2)/(3))x (log 3)/(log 2) (C) ((2)/(3))

Question

2^x

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.

Let's make this concrete. Suppose you have a function $f$ that depends on $g$ , and $g$ itself depends on $x$ :

y = f(g(x))

You want to know: if $x$ changes by a tiny amount, how much does $y$ change? The answer isn't just $f'(g(x))$ — because $g(x)$ itself changes when $x$ changes. You have to multiply the two rates:

How fast does $g$ change with respect to $x$ ? That's $g'(x)$ .
How fast does $f$ change with respect to its input $g$ ? That's $f'(g(x))$ .

The total effect is the product:

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

Note

In Leibniz notation, this looks even more natural: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ , where $u = g(x)$ . The $du$ 's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.

The Precise Statement

Chain Rule (single variable): If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $h(x) = f(g(x))$ is differentiable at $x$ , and

h'(x) = f'(g(x)) \cdot g'(x)

That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.

A Simple Example

Differentiate $h(x) = \sin(3x^2)$ .

Here $f(u) = \sin u$ and $g(x) = 3x^2$ . Then:

$f'(u) = \cos u$ , so $f'(g(x)) = \cos(3x^2)$
$g'(x) = 6x$

Multiply: $h'(x) = \cos(3x^2) \cdot 6x = 6x \cos(3x^2)$

Watch out

The most common mistake is forgetting to multiply by the inner derivative. Students often write $\frac{d}{dx} \sin(3x^2) = \cos(3x^2)$ and stop — that's wrong. The chain rule demands you also multiply by $6x$ .

Why It's Called a "Chain"

Think of a chain of links: $x \to g \to f$ . Each link has its own rate of change. To find the total rate from $x$ to $f$ , you multiply the rates of each link. If you had three functions — say $h(x) = f(g(k(x)))$ — you'd multiply three derivatives: …

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To find the derivative of $f(x)$ with respect to $g(x)$ , we compute the ratio of their individual derivatives with respect to $x$ . For $f(x) = 2^x$ and $g(x) = 3^x$ , this yields $\left(\frac{2}{3}\right)^x \frac{\log 2}{\log 3}$ .

When asked to find the derivative of one function with respect to another, say the derivative of $u = f(x)$ with respect to $v = g(x)$ , we are essentially looking for $\frac{du}{dv}$ . Since both $u$ and $v$ are functions of a common variable $x$ , we can use the chain rule.

The chain rule allows us to express $\frac{du}{dv}$ in terms of derivatives with respect to $x$ :

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

This means we need to:

Find the derivative of the first function ( $2^x$ ) with respect to $x$ .
Find the derivative of the second function ( $3^x$ ) with respect to $x$ .
Divide the first result by the second result.

Let's proceed with the steps.

Identify the functions:

Let $u = 2^x$ be the function whose derivative is required.

Let $v = 3^x$ be the function with respect to which we are differentiating.

We need to find $\frac{du}{dv}$ .
Differentiate $u$ with respect to $x$ :

We use the standard differentiation formula for exponential functions.
$a^x$
Differentiate $v$ with respect to $x$ :

Similarly, applying the formula to $v = 3^x$ :

$\frac{dv}{dx} = \frac{d}{dx}(3^x) = 3^x \log 3$ . …

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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 115 on this concept.

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

$\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}$ .

Let $u = x^2+ax+1$ , so $\frac{du}{dx} = 2x+a$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ : …

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

$\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ .

Let $u = x^2$ , so $\frac{du}{dx} = 2x$ .

By the chain rule, …

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

$\frac{d}{dx}\sqrt{\cot x} = \frac{-\csc^2 x}{2\sqrt{\cot x}}$ .

Let $u = \cot x$ , so $\frac{du}{dx} = -\csc^2 x$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ : …

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

$\frac{d}{dx}\cos\sqrt{x} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}$ .

Let $u = \sqrt{x}$ , so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ .

By the chain rule, …

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

Chain rule on $\cos(x^3)$ gives $-3x^2\sin x^3$ .

Let the inner function be $u=x^3$ , so $y=\cos u$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\log(\cos e^x)$
(a) $-\tan e^x$
(b) $e^x\tan e^x$
(c) $-e^x\tan e^x$
(d) $\tan e^x$
›Reveal solutionSolution

Using the chain rule twice, $\dfrac{d}{dx}\log(\cos e^x)=-e^x\tan e^x$ .

Let $y=\log(\cos e^x)$ . We differentiate using the chain rule, working from the outside in.

Step 1: Differentiate $\log(u)$ where $u=\cos e^x$ :

$\frac{dy}{dx}=\frac{1}{\cos e^x}\cdot\frac{d}{dx}(\cos e^x)$

Step 2: Differentiate $\cos(v)$ where $v=e^x$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
(a) $\frac{1}{1+x^2}$
(b) $\frac{2}{4+x}$
(c) $\frac{2}{1+x^2}$
(d) $\frac{2}{x+x^2}$
›Reveal solutionSolution

Put $x=\tan\phi$ so the expression simplifies to $\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$ , whose derivative is standard.

Let $x=\tan\phi$ . Then $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\phi}{1+\tan^2\phi} = \cos2\phi$ .

…

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CBSE 2026Set ANNUAL1 mark
$y = e^x + e^{x^2} + \ldots + e^{x^5}$
›Reveal solutionSolution

Differentiate each term $e^{x^n}$ using the chain rule: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$ .

$y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\cos 3x) =$
(a) $\sin 3x$
(b) $-3\sin 3x$
(c) $\cos 3x$
(d) $-3\cos 3x$
›Reveal solutionSolution

Differentiate $\cos(3x)$ using the chain rule: derivative of $\cos u$ is $-\sin u$ , times $\dfrac{du}{dx}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx} \tan^{-1}(x^2) =$
(a) $\dfrac{2x}{1+x^4}$
(b) $\dfrac{x}{1+x^2}$
(c) $\dfrac{x^3}{1+x^2}$
(d) None of these
›Reveal solutionSolution

Use $\dfrac{d}{dx}\tan^{-1}u = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}$ with $u=x^2$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx} e^{\sin x} =$
(a) $e^{\sin x} \cdot \cos x$
(b) $e^{\sin x}$
(c) $\cos x$
(d) None of these
›Reveal solutionSolution

Differentiate $e^{u}$ with $u=\sin x$ : derivative is $e^u \cdot \dfrac{du}{dx}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\log \sec x) =$
(a) $\tan x$
(b) $\cot x$
(c) $\csc x$
(d) None of these
›Reveal solutionSolution

Differentiate $\log(\sec x)$ using $\dfrac{d}{dx}\log u = \dfrac{u'}{u}$ with $u=\sec x$ .

…

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