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

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

The differential coefficient of $a^x + \log x \cdot \sin x$ is

If $y = \cos(\sin x^2)$,then $\frac{dy}{dx}$ at $x = \sqrt{\frac{\pi}{2}}$ is

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is

If $y = \frac{ax+b}{cx+d}$ and $\frac{dx}{dy} = \frac{ad-bc}{Py^2+Qy+R}$,then $P+Q+R =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module