The value of frac{d}{d x}left[log left(sin sq | vedclass

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

Question

(C) Let $y = \log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)$.Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot

Vedclass · Accepted Answer

$\frac{\sqrt{2} \cot \left(\frac{\sqrt{3}}{2}\right)}{8 \sqrt{3}}$

Vedclass · Answer

(C) Let $y = \log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)$.Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot \cos \sqrt{\frac{x^2+1}{x^2+2}} \cdot \frac{1}{2\sqrt{\frac{x^2+1}{x^2+2}}} \cdot \frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right)$.Calculating the derivative of the inner quotient: $\frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right) = \frac{2x(x^2+2) - 2x(x^2+1)}{(x^2+2)^2} = \frac{2x^3+4x-2x^3-2x}{(x^2+2)^2} = \frac{2x}{(x^2+2)^2}$.Substituting $x = \sqrt{2}$, we have $\frac{x^2+1}{x^2+2} = \frac{2+1}{2+2} = \frac{3}{4}$.Thus, $\frac{dy}{dx} = \cot \left( \sqrt{\frac{3}{4}} \right) \cdot \frac{1}{2 \sqrt{3/4}} \cdot \frac{2\sqrt{2}}{(2+2)^2} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{2 \cdot \frac{\sqrt{3}}{2}} \cdot \frac{2\sqrt{2}}{16} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{8} = \frac{\sqrt{2} \cot \left( \frac{\sqrt{3}}{2} \right)}{8 \sqrt{3}}$.

The value of $\frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right]$ when $x=\sqrt{2}$ is:

Similar Questions

If $y = (1 + x^2) \tan^{-1} x - x$,then $\frac{dy}{dx}$ is

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime}(a)=2$ and $f(a)=4$. Then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ equals ...... .

If $h(x) = \sqrt{4f(x) + 3g(x)}$,$f(1) = 4$,$g(1) = 3$,$f'(1) = 4$,and $g'(1) = 3$,then find $h'(1)$.

$10^{-x \tan x} \left[ \frac{d}{dx} (10^{x \tan x}) \right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module