The differential coefficient of tan^{-1}sqrt{ | vedclass

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

Question

(D) Let $y_1 = 	an^{-1}\sqrt{x}$ and $y_2 = \sqrt{x}$.Differentiating $y_1$ with respect to $x$ using the chain rule:$\frac{dy_1}{dx} = \frac{1}{1+(\

Vedclass · Accepted Answer

$\frac{1}{1+x}$

Vedclass · Answer

(D) Let $y_1 = 	an^{-1}\sqrt{x}$ and $y_2 = \sqrt{x}$.Differentiating $y_1$ with respect to $x$ using the chain rule:$\frac{dy_1}{dx} = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}$.Differentiating $y_2$ with respect to $x$:$\frac{dy_2}{dx} = \frac{1}{2\sqrt{x}}$.Now,the differential coefficient of $y_1$ with respect to $y_2$ is given by:$\frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{\frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} = \frac{1}{1+x}$.Thus,the correct option is $D$.

The differential coefficient of $\tan^{-1}\sqrt{x}$ with respect to $\sqrt{x}$ is

Similar Questions

The derivative of $\tan ^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]$ with respect to $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$

$\frac{d}{dx} \left\{ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right\} = $

If $y = \tan^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is

Let $f: R \rightarrow R$ be a continuous function. If $px+my+n=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$,then at $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module