If y = e^{m sin^{-1} x} and (1 - x^2) (frac{d | vedclass

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

Question

(C) Given,$y = e^{m \sin^{-1} x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{

Vedclass · Accepted Answer

$m^2$

Vedclass · Answer

(C) Given,$y = e^{m \sin^{-1} x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{-1} x) = y \cdot \frac{m}{\sqrt{1 - x^2}}$.Squaring both sides,we get:$(\frac{dy}{dx})^2 = \frac{m^2 y^2}{1 - x^2}$.Multiplying both sides by $(1 - x^2)$,we obtain:$(1 - x^2) (\frac{dy}{dx})^2 = m^2 y^2$.Comparing this with the given equation $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,we find that $A = m^2$.

If $y = e^{m \sin^{-1} x}$ and $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,then $A = . . . . . .$

Similar Questions

If $y = \log_2(\log_2(x))$,then $\frac{dy}{dx}$ is equal to

If $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$,then $\frac{dy}{dx} = \dots$

Find the derivative: $\frac{d}{dx}[(\log_e x)(\log_a x)]$

Differentiate the following with respect to $x$: $\log (\log x)$,where $x > 1$.

If $y=e^{\sin \left(\operatorname{cosec}^{-1} x\right)}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module