If f(x) = cos^{-1}left[ frac{1 - (log x)^2}{1 | vedclass

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

Question

(B) Given $f(x) = \cos^{-1}\left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$.Let $u = \log x$. Then the expression becomes $\cos^{-1}\left[ \frac{

Vedclass · Accepted Answer

$\frac{1}{e}$

Vedclass · Answer

(B) Given $f(x) = \cos^{-1}\left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} ight]$.Let $u = \log x$. Then the expression becomes $\cos^{-1}\left[ \frac{1 - u^2}{1 + u^2} ight]$.Using the trigonometric substitution $u = 	an 	heta$,we know that $\cos^{-1}\left[ \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight] = \cos^{-1}(\cos 2	heta) = 2	heta = 2	an^{-1} u$.Thus,$f(x) = 2	an^{-1}(\log x)$.Differentiating with respect to $x$ using the chain rule:$f'(x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{1}{x}$.Now,evaluating at $x = e$:$f'(e) = 2 \cdot \frac{1}{1 + (\log e)^2} \cdot \frac{1}{e}$.Since $\log e = 1$,we have $f'(e) = 2 \cdot \frac{1}{1 + 1^2} \cdot \frac{1}{e} = 2 \cdot \frac{1}{2} \cdot \frac{1}{e} = \frac{1}{e}$.

If $f(x) = \cos^{-1}\left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$,then the value of $f'(e)$ is:

Similar Questions

Let $f:R \to R$ be such that $f(1) = 3$ and $f'(1) = 6$. Then $\lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$ equals

Find the derivative of $\tan (2 x+3)$

Find the derivative of the following function: $\operatorname{cosec} x \cot x$.

If $5 f(x) + 3 f\left(\frac{1}{x}\right) = x + 2$ and $y = x f(x)$,then $\frac{dy}{dx}$ at $x = 1$ is equal to

If $f(x) = x^3 + e^{x/2}$ and $g(x) = f^{-1}(x)$,then the value of $g'(1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module