If y = log {left( frac{1 + x}{1 - x} right)^{ | vedclass

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

Question

(A) Given $y = \log {\left( \frac{1 + x}{1 - x} ight)^{1/4}} - \frac{1}{2}{	an ^{ - 1}}x$. Using properties of logarithms,$y = \frac{1}{4} [\log(1

Vedclass · Accepted Answer

$\frac{x^2}{1 - x^4}$

Vedclass · Answer

(A) Given $y = \log {\left( \frac{1 + x}{1 - x} ight)^{1/4}} - \frac{1}{2}{	an ^{ - 1}}x$. Using properties of logarithms,$y = \frac{1}{4} [\log(1 + x) - \log(1 - x)] - \frac{1}{2}{	an ^{ - 1}}x$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} - \frac{1}{1 - x}(-1) ight] - \frac{1}{2} \left( \frac{1}{1 + x^2} ight)$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} + \frac{1}{1 - x} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1 - x + 1 + x}{(1 + x)(1 - x)} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{2}{1 - x^2} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{2(1 - x^2)} - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{(1 + x^2) - (1 - x^2)}{(1 - x^2)(1 + x^2)} ight]$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{2x^2}{1 - x^4} ight] = \frac{x^2}{1 - x^4}$.

If $y = \log {\left( \frac{1 + x}{1 - x} \right)^{1/4}} - \frac{1}{2}{\tan ^{ - 1}}x,$ then $\frac{dy}{dx} = $

Similar Questions

For some constants $a$ and $b$,find the derivative of $(x-a)(x-b)$.

If $y = \sin (\sqrt {\sin x + \cos x} )$,then $\frac{dy}{dx} = $

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

$\frac{d}{dx}(e^x \log \sin 2x) = $

Let $f(x + y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$,where $g(x)$ is continuous. Then $f'(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module