If y = tan(cos^{-1} x),then frac{dy}{dx} is e | vedclass

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

Question

(D) Given $y = 	an(\cos^{-1} x)$.Let $\cos^{-1} x = 	heta$,then $\cos 	heta = x$.We know that $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos \

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $y = 	an(\cos^{-1} x)$.Let $\cos^{-1} x = 	heta$,then $\cos 	heta = x$.We know that $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos 	heta} = \frac{\sqrt{1 - x^2}}{x}$.So,$y = \frac{\sqrt{1 - x^2}}{x}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1 - x^2}) - \sqrt{1 - x^2} \cdot \frac{d}{dx}(x)}{x^2}$$\frac{dy}{dx} = \frac{x \cdot \frac{-2x}{2\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$$\frac{dy}{dx} = \frac{\frac{-x^2}{\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$$\frac{dy}{dx} = \frac{-x^2 - (1 - x^2)}{x^2 \sqrt{1 - x^2}} = \frac{-1}{x^2 \sqrt{1 - x^2}}$.

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

Similar Questions

Find the derivative: $\frac{d}{dx} \sqrt{x \sin x}$

If $f$ is differentiable,$f(x+y)=f(x) f(y)$ for all $x, y \in R$,$f(3)=3$,and $f^{\prime}(0)=11$,then $f^{\prime}(3)$ is equal to:

The values of $x$,at which the first derivative of the function $(\sqrt{x} + \frac{1}{\sqrt{x}})^2$ with respect to $x$ is $\frac{3}{4}$,are

$\frac{d}{dx}(e^{x\sin x}) = $

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module