(a) If $y = x^{\frac{1}{x}}$, then find $\frac{dy}{dx}$ at $x = 1$. OR (b) If $x = a \sin 2t$, $y = a(\cos 2t + \log \tan t)$, then find $\frac{dy}{dx}$.

(a) If y = x(1)/(x), then find (dy)/(dx) at x = 1. OR (b) If x = a sin 2t, y = a(cos 2t + log tan t), then fin

Question

y = x^{\frac{1}{x}}

(OR)

(b) If

x = a \sin 2t

y = a(\cos 2t + \log \tan t)

, then find

\frac{dy}{dx}

CBSECBSE Class XII Board 2023Subjective· 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 — 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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$\dfrac{dy}{dx}\big|_{x=1}=1$ .
$\dfrac{dy}{dx}=\cot 2t$ .

Part (a): $y=x^{1/x}$ , find $\frac{dy}{dx}$ at $x=1$

Both base and exponent are variable, so use logarithmic differentiation.

1. $\ln y=\ln\!\big(x^{1/x}\big)=\dfrac{1}{x}\ln x.$

2. Differentiate (product rule on the right, chain rule on the left):

\frac1y\frac{dy}{dx}=\Big(-\frac1{x^2}\Big)\ln x+\frac1x\cdot\frac1x=\frac{1-\ln x}{x^2}.

3. Isolate and substitute $y=x^{1/x}$ :

\frac{dy}{dx}=x^{1/x}\,\frac{1-\ln x}{x^2}.

4. Evaluate at $x=1$ ( $x^{1/x}=1,\ \ln1=0$ ): …

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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 218 on this concept.

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
$\frac{d}{dx}\sqrt{x^2+ax+1} =$
(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
$\frac{d}{dx}(\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x}) =$
(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
$\frac{d}{dx}(\cos\sqrt{x}) =$
(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
$\frac{d}{dx}\left\{\lim_{x\to 0}\frac{x^5-a^5}{x-a}\right\} =$
(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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Q.(a) If y=x1xy = x^{\frac{1}{x}}y=xx1​, then find dydx\frac{dy}{dx}dxdy​ at x=1x = 1x=1.

Part (a): y=x1/xy=x^{1/x}y=x1/x, find dydx\frac{dy}{dx}dxdy​ at x=1x=1x=1

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Q.(a) If $y = x^{\frac{1}{x}}$ , then find $\frac{dy}{dx}$ at $x = 1$ .

Part (a): $y=x^{1/x}$ , find $\frac{dy}{dx}$ at $x=1$

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