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

$\frac{d}{dx} \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = $

If $t = \frac{v^2}{2}$,then $\left( - \frac{df}{dt} \right)$ is equal to,(where $f$ is acceleration)

If $e^{x}=y+\sqrt{y^2-1}$, then $\frac{d y}{d x}=$

Let $y = \sqrt{p x^3 y}$. If $\frac{dy}{dx} = \frac{3}{2}$ when $y = 1$, then the value of $p$ is equal to

Find the derivative: $\frac{d}{dx}[\cos((1 - x^2)^2)] = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module