If f(x) = sin^{-1}left(frac{2 cdot 3^x}{1 + 9 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.$f(x) = \sin^{-1}\left

Vedclass · Accepted Answer

$\sqrt{3} \ln(\sqrt{3})$

Vedclass · Answer

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.$f(x) = \sin^{-1}\left(\frac{2 	an 	heta}{1 + 	an^2 	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(3^x)$.Differentiating with respect to $x$:$f'(x) = 2 \cdot \frac{1}{1 + (3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2}{1 + 9^x} \cdot 3^x \ln 3$.Now,evaluate at $x = -\frac{1}{2}$:$f'(-\frac{1}{2}) = \frac{2 \cdot 3^{-1/2}}{1 + 9^{-1/2}} \ln 3 = \frac{2 / \sqrt{3}}{1 + 1/3} \ln 3 = \frac{2 / \sqrt{3}}{4/3} \ln 3 = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} \ln 3 = \frac{\sqrt{3}}{2} \ln 3$.Since $\ln 3 = 2 \ln \sqrt{3}$,we have $f'(-\frac{1}{2}) = \frac{\sqrt{3}}{2} \cdot 2 \ln \sqrt{3} = \sqrt{3} \ln \sqrt{3}$.

If $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right)$,then $f'(-\frac{1}{2})$ equals

Similar Questions

If $y = \tan^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} \right]$, then $\frac{dy}{dx}$ is equal to

The derivative of $\sin ^{-1}\left(3 x-4 x^3\right)$ with respect to $x$ for $\frac{1}{2} < x < 1$ is:

$\frac{d}{dx} \tan^{-1} \left( \frac{4\sqrt{x}}{1 - 4x} \right) = $

$\frac{d}{dx} \left( \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}} \right) = $

$\frac{d}{dx} \tan^{-1} \left[ \frac{\cos x - \sin x}{\cos x + \sin x} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module