If y = x^(sin x) + (sin x)x then find dy/dx. **OR** If x = (1 − t²)/(1 + t²), y = 2t/(1 + t²) then prove that

Question of 281

Q. If y = x^(sin x) + (sin x)^x then find dy/dx. OR If x = (1 - t²)/(1 + t²), y = 2t/(1 + t²) then prove that dy/dx + x/y = 0.

Punjab PsebPSEB Punjab Class 12 Board 2025Subjective· 4mImportance★★★★★

0% · 0/281 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 →

Since both terms have a variable base AND a variable exponent, differentiate each term separately using logarithmic differentiation, then add the results.

Let $u = x^{\sin x}$ and $v = (\sin x)^x$ , so $y = u+v$ .

For $u$ : take logs: $\ln u = \sin x\ln x$ . Differentiating implicitly:

$\frac{1}{u}\frac{du}{dx} = \cos x\ln x + \sin x\cdot\frac1x \implies \frac{du}{dx} = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right).$

For $v$ : take logs: $\ln v = x\ln(\sin x)$ . Differentiating implicitly:

$\frac{1}{v}\frac{dv}{dx} = \ln(\sin x) + x\cdot\frac{\cos x}{\sin x} \implies \frac{dv}{dx} = (\sin x)^x\big(\ln(\sin x) + x\cot x\big).$

Adding:

$\frac{dy}{dx} = x^{\sin x}\!\left(\cos x\ln x + \frac{\sin x}{x}\right) + (\sin x)^x\big(\ln(\sin x) + x\cot x\big).$

…

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.

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

$\frac{dy}{dx} = x^x(\log x + 1)$ .

Take logs of $y = x^x$ :

$\log y = x\log x.$

Differentiate implicitly:

$\frac{1}{y}\frac{dy}{dx} = \log x + x\cdot\frac{1}{x} = \log x + 1.$

So …

Start your 14-day free trial to see the full solution →
CBSE 2026Set SEM31 markMCQ
$x^m y^n = (x + y)^{m+n}$
(a) $0$
(b) $\dfrac{y}{x}$
(c) $\dfrac{x + y}{xy}$
(d) $xy$
›Reveal solutionSolution

Use logarithmic differentiation on $x^m y^n=(x+y)^{m+n}$ ; the cross terms cancel and $\dfrac{dy}{dx}=\dfrac{y}{x}$ .

Logarithmic differentiation of an implicit relation is a standard NCERT/CBSE Class 12 continuity and differentiability method.

Take natural logs of both sides:

$m\ln x + n\ln y = (m+n)\ln(x+y).$

Differentiate with respect to $x$ :

$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = (m+n)\cdot\frac{1+\tfrac{dy}{dx}}{x+y}.$

Multiply out and collect. Writing $y'=\tfrac{dy}{dx}$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The derivative of x^(2x) with respect to x is ......................$
(a) x^(2x−1)
(b) 2x^(2x) log x
(c) 2x^(2x)(1 + log x)
(d) 2x^(2x)(1 − 2 log x)
›Reveal solutionSolution

When the variable appears in both the base and the exponent, take logarithms first (logarithmic differentiation), then differentiate implicitly.

Given: $y = x^{2x}$

Step 1 — take log of both sides:

$\ln y = 2x \ln x$

Step 2 — differentiate both sides w.r.t. $x$ (product rule on the right):

$\frac{1}{y}\frac{dy}{dx} = 2\ln x + 2x \cdot \frac{1}{x} = 2\ln x + 2$

…

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

←Previous chapterCh 4Determinants

Q.If y = x^(sin x) + (sin x)^x then find dy/dx. OR If x = (1 − t²)/(1 + t²), y = 2t/(1 + t²) then prove that dy/dx + x/y = 0.

Concept understanding — Logarithmic Differentiation

Q.If y = x^(sin x) + (sin x)^x then find dy/dx. OR If x = (1 − t²)/(1 + t²), y = 2t/(1 + t²) then prove that dy/dx + x/y = 0.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days