If y=e^{sin ^{-1} x},then left(1-x^2right) y_ | vedclass

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

Question

(C) Given,$y=e^{\sin ^{-1} x}$.Differentiating with respect to $x$:$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}

Vedclass · Accepted Answer

y

Vedclass · Answer

(C) Given,$y=e^{\sin ^{-1} x}$.Differentiating with respect to $x$:$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$$y_1 = \frac{y}{\sqrt{1-x^2}}$$y_1 \sqrt{1-x^2} = y$Differentiating both sides with respect to $x$ using the product rule:$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = y_1$$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = y_1$Multiplying both sides by $\sqrt{1-x^2}$:$y_2 (1-x^2) - x y_1 = y_1 \sqrt{1-x^2}$Since $y_1 \sqrt{1-x^2} = y$,we have:$(1-x^2) y_2 - x y_1 = y$Thus,option $(c)$ is correct.

If $y=e^{\sin ^{-1} x}$,then $\left(1-x^2\right) y_2-x y_1=$

Similar Questions

If $ y = (\tan^{-1} x)^2 $,then $ (x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 $ is equal to

If $f(x)=b \cdot e^{a x}+a \cdot e^{b x}$,then $f^{\prime \prime}(0)=$

If $y = a^x \cdot b^{2x-1}$,then $\frac{d^2 y}{d x^2}$ is equal to

If ${y^2} = a{x^2} + bx + c$,then ${y^3}\frac{{{d^2}y}}{{d{x^2}}}$ is

$\frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module