If y=tan ^{-1}left(frac{2-3 sin x}{3-2 sin x} | vedclass

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

Question

(B) Given,$y=	an ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Using the chain rule,$\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x

Vedclass · Accepted Answer

$\frac{-5 \cos x}{13 \sin ^2 x-24 \sin x+13}$

Vedclass · Answer

(B) Given,$y=	an ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Using the chain rule,$\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x}ight)^2} 	imes \frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Simplify the denominator: $1+\left(\frac{2-3 \sin x}{3-2 \sin x}ight)^2 = \frac{(3-2 \sin x)^2 + (2-3 \sin x)^2}{(3-2 \sin x)^2} = \frac{9+4 \sin^2 x - 12 \sin x + 4 + 9 \sin^2 x - 12 \sin x}{(3-2 \sin x)^2} = \frac{13 \sin^2 x - 24 \sin x + 13}{(3-2 \sin x)^2}$.Now,differentiate the inner function using the quotient rule: $\frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}ight) = \frac{(3-2 \sin x)(-3 \cos x) - (2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}$.$= \frac{-9 \cos x + 6 \sin x \cos x + 4 \cos x - 6 \sin x \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{(3-2 \sin x)^2}$.Combining these,$\frac{d y}{d x} = \frac{(3-2 \sin x)^2}{13 \sin^2 x - 24 \sin x + 13} 	imes \frac{-5 \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{13 \sin^2 x - 24 \sin x + 13}$.

If $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$,then $\frac{d y}{d x}=$

Similar Questions

If $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$,then $\frac{dy}{dx} = $

The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ is

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

The differential coefficient of ${\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ with respect to $\sqrt {1 - {x^2}} $ at $x = \frac{1}{2}$ is:

$\frac{d}{dx}\left( \tan^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module