((x³-5)⁵)/((x³+3)³)

EXERCISE 1.1 · Q28

\dfrac{(x^3-5)^5}{(x^3+3)^3}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

10% · 28/293 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

Let $y=\dfrac{(x^3-5)^5}{(x^3+3)^3}$ . Take natural logs of both sides: $\ln y=5\ln(x^3-5)-3\ln(x^3+3)$ .

Step 1 — differentiate both sides w.r.t. $x$ (implicit differentiation on the left): $\dfrac{1}{y}\dfrac{dy}{dx}=5\cdot\dfrac{3x^2}{x^3-5}-3\cdot\dfrac{3x^2}{x^3+3}=\dfrac{15x^2}{x^3-5}-\dfrac{9x^2}{x^3+3}$ .

Step 2 — multiply both sides by $y$ : $\dfrac{dy}{dx}=\dfrac{(x^3-5)^5}{(x^3+3)^3}\left[\dfrac{15x^2}{x^3-5}-\dfrac{9x^2}{x^3+3}\right]=\dfrac{15x^2(x^3-5)^4}{(x^3+3)^3}-\dfrac{9x^2(x^3-5)^5}{(x^3+3)^4}$ . …

Start your 14-day free trial to unlock the full solution →

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

Start your 14-day free trial to see the full solution →
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, …

Start your 14-day free trial to see the full solution →
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 $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ : …

Start your 14-day free trial to see the full solution →
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, …

Start your 14-day free trial to see the full solution →
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 $y=\cos u$ .

…

Start your 14-day free trial to see the full solution →
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 $v=e^x$ : …

Start your 14-day free trial to see the full solution →
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 $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\phi}{1+\tan^2\phi} = \cos2\phi$ .

…

Start your 14-day free trial to see the full solution →
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: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$ .

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

…

Start your 14-day free trial to see the full solution →
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 $\dfrac{du}{dx}$ .

…

Start your 14-day free trial to see the full solution →
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 $u=x^2$ .

…

Start your 14-day free trial to see the full solution →
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 $e^u \cdot \dfrac{du}{dx}$ .

…

Start your 14-day free trial to see the full solution →
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 $u=\sec x$ .

…

Start your 14-day free trial to see the full solution →

←Previous questionQ27 — dfracx√(7-3x)→Next questionQ29 — (1+sin² x)²(1+cos² x)³

Q.(x3−5)5(x3+3)3\dfrac{(x^3-5)^5}{(x^3+3)^3}(x3+3)3(x3−5)5​

🎓Unlock everything free for 14 days

Q. $\dfrac{(x^3-5)^5}{(x^3+3)^3}$

Unlock everything free for 14 days