If m sin ^{-1} x = log _{e} y,then (1 - x^{2} | vedclass

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

Question

(A) Given,$m \sin ^{-1} x = \log _{e} y$ $\Rightarrow y = e^{m \sin ^{-1} x}$ Differentiating with respect to $x$,we get: $y' = e^{m \sin ^{-1} x} 	i

Vedclass · Accepted Answer

$m^{2} y$

Vedclass · Answer

(A) Given,$m \sin ^{-1} x = \log _{e} y$ $\Rightarrow y = e^{m \sin ^{-1} x}$ Differentiating with respect to $x$,we get: $y' = e^{m \sin ^{-1} x} 	imes \frac{m}{\sqrt{1 - x^{2}}}$ $\Rightarrow \sqrt{1 - x^{2}} \cdot y' = m y$ Squaring both sides,we get: $(1 - x^{2}) (y')^{2} = m^{2} y^{2}$ Differentiating again with respect to $x$: $(1 - x^{2}) \cdot 2 y' \cdot y'' + (y')^{2} (-2 x) = m^{2} \cdot 2 y y' $ Dividing both sides by $2 y'$,we get: $(1 - x^{2}) y'' - x y' = m^{2} y$

If $m \sin ^{-1} x = \log _{e} y$,then $(1 - x^{2}) y'' - x y'$ is equal to

Similar Questions

If $x^x y^y=e^e$,then $\left(\frac{d^2 y}{d x^2}\right)_{(e, e)}=$

If $y = x + e^x$,then $\frac{d^2x}{dy^2}$ is:

Find the second order derivative of the function $\tan^{-1} x$.

If $f(x) = e^{(x+1)^n}; (n \in N)$,then the value of $n$ for which $f''(1) = 67(2^n e^{2^n})$ is

If $y=3 e^{5 x}+5 e^{3 x}$,then $\frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}=$ (in $y$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module