If y = log sqrt{tan x},then the value of frac | vedclass

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

Question

(A) Given $y = \log \sqrt{	an x} = \frac{1}{2} \log(	an x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $y = \log \sqrt{	an x} = \frac{1}{2} \log(	an x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} 	imes \frac{1}{	an x} 	imes \sec^2 x$.Using the identities $	an x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we get:$\frac{dy}{dx} = \frac{1}{2} 	imes \frac{\cos x}{\sin x} 	imes \frac{1}{\cos^2 x} = \frac{1}{2 \sin x \cos x} = \frac{1}{\sin(2x)}$.Now,substitute $x = \frac{\pi}{4}$:$\left(\frac{dy}{dx}ight)_{x=\frac{\pi}{4}} = \frac{1}{\sin(2 	imes \frac{\pi}{4})} = \frac{1}{\sin(\frac{\pi}{2})} = \frac{1}{1} = 1$.

If $y = \log \sqrt{\tan x}$,then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$ is

Similar Questions

$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $

The derivative of $f(x) = 3|2 + x|$ at the point $x_0 = -3$ is

Differentiate the following with respect to $x$: $e^{x^{3}}$

Find the derivative of the function: $(x^{2}+1) \cos x$. (Assume that $a, b, c, d, p, q, r, s$ are fixed non-zero constants and $m, n$ are integers.)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module