If y = frac{1}{sqrt{a^2 - b^2}} cos^{-1} left | vedclass

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

Question

(A) Let $	heta = a + b \cos(x - \alpha)$. The given expression is $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left( \frac{a \cos(x - \alpha) + b}{	he

Vedclass · Accepted Answer

$\frac{1}{a + b \cos(x - \alpha)}$

Vedclass · Answer

(A) Let $	heta = a + b \cos(x - \alpha)$. The given expression is $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left( \frac{a \cos(x - \alpha) + b}{	heta} ight)$.Taking $\cos$ on both sides: $\cos(y \sqrt{a^2 - b^2}) = \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)}$.Differentiating both sides with respect to $x$:$-\sqrt{a^2 - b^2} \sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{d}{dx} \left( \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} ight)$.Using the quotient rule,the derivative of the right side is $\frac{-a \sin(x - \alpha)(a + b \cos(x - \alpha)) - (a \cos(x - \alpha) + b)(-b \sin(x - \alpha))}{(a + b \cos(x - \alpha))^2} = \frac{-(a^2 - b^2) \sin(x - \alpha)}{	heta^2}$.Thus,$-\sqrt{a^2 - b^2} \sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{-(a^2 - b^2) \sin(x - \alpha)}{	heta^2}$.Simplifying,$\sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta^2}$.Since $\sin(y \sqrt{a^2 - b^2}) = \sqrt{1 - \cos^2(y \sqrt{a^2 - b^2})} = \sqrt{1 - \left( \frac{a \cos(x - \alpha) + b}{	heta} ight)^2} = \frac{\sqrt{	heta^2 - (a \cos(x - \alpha) + b)^2}}{	heta} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta}$.Substituting this back: $\left( \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta} ight) \frac{dy}{dx} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta^2}$.Therefore,$\frac{dy}{dx} = \frac{1}{	heta} = \frac{1}{a + b \cos(x - \alpha)}$.

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} = $

Similar Questions

If $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$,then $\frac{d y}{d x}=$

If $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$,then $f^{\prime}(x) = $

If $y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$, then find the value of $y'(1)$.

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

The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $\cos ^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module