If tan u=sqrt{frac{1-x}{1+x}} and cos v=4 x^{ | vedclass

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

Question

(A) Given $	an u=\sqrt{\frac{1-x}{1+x}}$.Let $x=\cos 	heta$,then $	heta=\cos ^{-1} x$.Substituting $x$,we get $	an u=\sqrt{\frac{1-\cos 	heta}{1+

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) Given $	an u=\sqrt{\frac{1-x}{1+x}}$.Let $x=\cos 	heta$,then $	heta=\cos ^{-1} x$.Substituting $x$,we get $	an u=\sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}}=\sqrt{\frac{2 \sin ^{2} (	heta/2)}{2 \cos ^{2} (	heta/2)}}=	an (	heta/2)$.Thus,$u=\frac{	heta}{2}=\frac{1}{2} \cos ^{-1} x$.Differentiating with respect to $x$,we get $\frac{du}{dx}=\frac{1}{2} \left(-\frac{1}{\sqrt{1-x^{2}}}ight) = -\frac{1}{2 \sqrt{1-x^{2}}}$.Given $\cos v=4 x^{3}-3 x$.Substituting $x=\cos 	heta$,we get $\cos v=4 \cos ^{3} 	heta-3 \cos 	heta = \cos 3 	heta$.Thus,$v=3 	heta = 3 \cos ^{-1} x$.Differentiating with respect to $x$,we get $\frac{dv}{dx}=3 \left(-\frac{1}{\sqrt{1-x^{2}}}ight) = -\frac{3}{\sqrt{1-x^{2}}}$.Finally,$\frac{du}{dv}=\frac{du/dx}{dv/dx} = \frac{-1/(2 \sqrt{1-x^{2}})}{-3/\sqrt{1-x^{2}}} = \frac{1}{6}$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

Similar Questions

The equation of the normal to the curve $x = \theta + \sin \theta, y = 1 + \cos \theta$ at $\theta = \frac{\pi}{2}$ is

The slope of the tangent to the curve $x = t^2 + 3t - 8$,$y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to the $X$-axis,then the value of $t$ is

The derivative of $\cot ^{-1} x$ with respect to $\log (1+x^{2})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module