If y=log_{sin x} tan x,then left(frac{dy}{dx} | vedclass

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

Question

(C) Given $y = \log_{\sin x} 	an x$. Using the change of base formula,we can write $y = \frac{\log 	an x}{\log \sin x}$.Applying the quotient rule $

Vedclass · Accepted Answer

$\frac{-4}{\log 2}$

Vedclass · Answer

(C) Given $y = \log_{\sin x} 	an x$. Using the change of base formula,we can write $y = \frac{\log 	an x}{\log \sin x}$.Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$,where $u = \log 	an x$ and $v = \log \sin x$:$u' = \frac{1}{	an x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.$v' = \frac{1}{\sin x} \cdot \cos x = \cot x$.Thus,$\frac{dy}{dx} = \frac{(\log \sin x)(2 \csc 2x) - (\log 	an x)(\cot x)}{(\log \sin x)^2}$.At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$	an x = 1$,$\log 	an x = 0$,$\csc 2x = \csc \frac{\pi}{2} = 1$,and $\cot x = 1$.Substituting these values:$\left(\frac{dy}{dx}ight)_{x=\frac{\pi}{4}} = \frac{(\log \frac{1}{\sqrt{2}})(2 \cdot 1) - (0)(1)}{(\log \frac{1}{\sqrt{2}})^2} = \frac{2 \log(2^{-1/2})}{(\log 2^{-1/2})^2} = \frac{2 \cdot (-\frac{1}{2} \log 2)}{(-\frac{1}{2} \log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

If $y=\log_{\sin x} \tan x$,then $\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}$ has the value

Similar Questions

If $y = \log_{\cos x} \sin x$,then $\frac{dy}{dx}$ is equal to

$\frac{d}{dx} \left[ \log \sqrt{\sin \sqrt{e^x}} \right] = $

Let $f(x)=e^x$, $g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$, then $\frac{h'(x)}{h(x)}$ is equal to

If $y = \log_{10}(x^2)$,then $\frac{dy}{dx}$ is equal to

If $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$,then the value of $f'(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module