If y = log_{sin x} (tan x),then left( frac{dy | 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 $y = \frac{\ln(	an x)}{\ln(\sin x)}$.Applying the quotient rule

Vedclass · Accepted Answer

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

Vedclass · Answer

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

If $y = \log_{\sin x} (\tan x)$,then $\left( \frac{dy}{dx} \right)_{\pi/4}$ is equal to

Similar Questions

If $y = \log(x^x)$,then $\frac{dy}{dx} = $

If $y = \log_2(\log_2 x)$,then $\frac{dy}{dx}$ is equal to

If $y = 3^{x^2}$,then $\frac{dy}{dx}$ is equal to

The value of $\log _{e} 2 \cdot \frac{d}{dx}(\log _{\cos x} \operatorname{cosec} x)$ at $x=\frac{\pi}{4}$ is.

$\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\} \right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module