y = log left( frac{sqrt{x^2+1}-x}{sqrt{x^2+1} | vedclass

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

Question

(D) Given $y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right)$.Rationalizing the denominator inside the logarithm:$y = \log \left( \frac{(\

Vedclass · Accepted Answer

$\frac{-2}{\sqrt{1+x^2}}$

Vedclass · Answer

(D) Given $y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right)$.Rationalizing the denominator inside the logarithm:$y = \log \left( \frac{(\sqrt{x^2+1}-x)^2}{(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)} \right) = \log \left( \frac{(\sqrt{x^2+1}-x)^2}{(x^2+1)-x^2} \right) = \log (\sqrt{x^2+1}-x)^2$.Using the property $\log(a^b) = b \log a$,we get $y = 2 \log(\sqrt{x^2+1}-x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{x^2+1}-x} \cdot \left( \frac{1}{2\sqrt{x^2+1}} \cdot 2x - 1 \right)$.$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{x^2+1}-x} \cdot \left( \frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}} \right)$.$\frac{dy}{dx} = 2 \cdot \frac{-( \sqrt{x^2+1}-x )}{\sqrt{x^2+1}(\sqrt{x^2+1}-x)} = \frac{-2}{\sqrt{x^2+1}}$.

$y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right) \Rightarrow \frac{dy}{dx} = $

Similar Questions

$\frac{d}{dx}(\log_5 x^2) = $ . . . . . .

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

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

If $f(x) = e^{g(x)}$ and $g(x) = \int_{2}^{x} \frac{t}{1 + t^4} \, dt$,then $f'(2)$ has the value equal to:

The differential coefficient of the function $\log_e \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$ with respect to $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module