If y=log left[tan sqrt{frac{2^x-1}{2^x+1}}rig | vedclass

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

Question

(C) Given $y=\log \left[	an \sqrt{\frac{2^x-1}{2^x+1}}ight]$.Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$,$v = \frac{2-1}{2+1} = \frac{1}{3}$.Using the

Vedclass · Accepted Answer

$\frac{4 \sqrt{3} \log 2}{9 \sin \left(\frac{2}{\sqrt{3}}\right)}$

Vedclass · Answer

(C) Given $y=\log \left[	an \sqrt{\frac{2^x-1}{2^x+1}}ight]$.Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$,$v = \frac{2-1}{2+1} = \frac{1}{3}$.Using the quotient rule,$\frac{dv}{dx} = \frac{(2^x+1)(2^x \log 2) - (2^x-1)(2^x \log 2)}{(2^x+1)^2} = \frac{2^x \log 2 (2^x+1-2^x+1)}{(2^x+1)^2} = \frac{2 \cdot 2^x \log 2}{(2^x+1)^2} = \frac{2^{x+1} \log 2}{(2^x+1)^2}$.At $x=1$,$\left(\frac{dv}{dx}ight)_{x=1} = \frac{2^2 \log 2}{(2+1)^2} = \frac{4 \log 2}{9}$.Now,$y = \log(	an \sqrt{v})$. Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{	an \sqrt{v}} \cdot \sec^2 \sqrt{v} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}$.Using $\frac{1}{	an \sqrt{v}} \cdot \sec^2 \sqrt{v} = \frac{\cos \sqrt{v}}{\sin \sqrt{v}} \cdot \frac{1}{\cos^2 \sqrt{v}} = \frac{1}{\sin \sqrt{v} \cos \sqrt{v}} = \frac{2}{\sin(2\sqrt{v})}$.So,$\frac{dy}{dx} = \frac{2}{\sin(2\sqrt{v})} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx} = \frac{1}{\sin(2\sqrt{v}) \cdot \sqrt{v}} \cdot \frac{dv}{dx}$.At $x=1$,$v = \frac{1}{3}$,so $\sqrt{v} = \frac{1}{\sqrt{3}}$.$\left(\frac{dy}{dx}ight)_{x=1} = \frac{1}{\sin(2/\sqrt{3}) \cdot (1/\sqrt{3})} \cdot \frac{4 \log 2}{9} = \frac{\sqrt{3} \cdot 4 \log 2}{9 \sin(2/\sqrt{3})} = \frac{4 \sqrt{3} \log 2}{9 \sin(2/\sqrt{3})}$.

If $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right], x>0$,then $\left(\frac{d y}{d x}\right)_{x=1}=$

Similar Questions

If $f(x) = x + 2,$ then $f'(f(x))$ at $x = 4$ is

If $f(x) = \log_{x^2} (\ln x)$,then $f'(e)$ is equal to . . . . . .

If $y = \frac{(a - x)\sqrt{a - x} - (b - x)\sqrt{x - b}}{\sqrt{a - x} + \sqrt{x - b}}$,then $\frac{dy}{dx}$ wherever it is defined is equal to:

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,where $x \in R$. Then $f'(10)$ is equal to:

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$,then $\left(\frac{dy}{dx}\right)_{x=0}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module