If f(x) = sin^{-1}left[frac{2^{x+1}}{1+4^x}ri | vedclass

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

Question

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

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(A) Given $f(x) = \sin^{-1}\left[\frac{2 \cdot 2^x}{1 + (2^x)^2}ight]$.Let $2^x = 	an 	heta$,then $	heta = 	an^{-1}(2^x)$.Substituting this into the function:$f(x) = \sin^{-1}\left[\frac{2 	an 	heta}{1 + 	an^2 	heta}ight]$Using the identity $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we get:$f(x) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(2^x)$.Now,differentiating with respect to $x$:$f'(x) = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x)$$f'(x) = \frac{2}{1 + 4^x} \cdot 2^x \log_e 2$.Evaluating at $x = 0$:$f'(0) = \frac{2}{1 + 4^0} \cdot 2^0 \log_e 2 = \frac{2}{1 + 1} \cdot 1 \cdot \log_e 2 = \frac{2}{2} \log_e 2 = \log_e 2$.

If $f(x) = \sin^{-1}\left[\frac{2^{x+1}}{1+4^x}\right]$,then $f'(0) = $

Similar Questions

If $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$,then $\frac{d u}{d v}$ at $x=0$ is

If $y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right)$,where $x^2 \le 1$. Then find $\frac{dy}{dx}$.

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ is:

Differentiate the following with respect to $x$: $\cos ^{-1}(\sin x)$

If $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$,then $\frac{dy}{dx}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module