If y = frac{sin^{-1} x}{sqrt{1 - x^2}},then ( | vedclass

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

Question

(B) Given $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$.Applying the quotient rule for differentiation,$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{d}

Vedclass · Accepted Answer

$1 + xy$

Vedclass · Answer

(B) Given $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$.Applying the quotient rule for differentiation,$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{d}{dx}(\sin^{-1} x) - \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1 - x^2})}{1 - x^2}$.Since $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$,we have:$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1} x \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{1 - x^2}$.$\frac{dy}{dx} = \frac{1 + \frac{x \sin^{-1} x}{\sqrt{1 - x^2}}}{1 - x^2}$.Multiplying both sides by $(1 - x^2)$,we get:$(1 - x^2)\frac{dy}{dx} = 1 + x \left( \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \right)$.Since $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$,we substitute $y$ into the expression:$(1 - x^2)\frac{dy}{dx} = 1 + xy$.

If $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$,then $(1 - x^2)\frac{dy}{dx}$ is equal to

Similar Questions

If $y=5 \sin x+6 \cos x$,then $y^2+(y_1)^2=$ . . . . . .

If $y = (1 + x^2)\tan^{-1}x - x,$ then $\frac{dy}{dx} = $

Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

If $y = \sqrt{\sin \sqrt{x}}$,then $\frac{dy}{dx} = $

Find the derivative of the following function: $\frac{4x + 5 \sin x}{3x + 7 \cos x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module