If y = x + e^x,then at x = 1,frac{d^2x}{dy^2} | vedclass

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

Question

(B) Given $y = x + e^x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + e^x$.Therefore,$\frac{dx}{dy} = \frac{1}{1 + e^x}$.Now,differe

Vedclass · Accepted Answer

$\frac{-e}{(1+e)^3}$

Vedclass · Answer

(B) Given $y = x + e^x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + e^x$.Therefore,$\frac{dx}{dy} = \frac{1}{1 + e^x}$.Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the chain rule:$\frac{d^2x}{dy^2} = \frac{d}{dx}\left(\frac{1}{1 + e^x}\right) \cdot \frac{dx}{dy}$$= \frac{-e^x}{(1 + e^x)^2} \cdot \frac{1}{1 + e^x} = \frac{-e^x}{(1 + e^x)^3}$.At $x = 1$,we have:$\left. \frac{d^2x}{dy^2} \right|_{x=1} = \frac{-e^1}{(1 + e^1)^3} = \frac{-e}{(1 + e)^3}$.

If $y = x + e^x$,then at $x = 1$,$\frac{d^2x}{dy^2}$ is equal to

Similar Questions

If $\sqrt{x+y}-\sqrt{x-y}=c$,then $\frac{d^2 y}{d x^2}=$

If $y=a \cos (\log x)+b \sin (\log x)$,where $a, b$ are parameters,then $x^2 y^{\prime \prime}+x y^{\prime}$ is equal to

If $y=e^{ax}(\cos bx+\sin bx)$ satisfies the equation $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$,then $L+bK=$

If $y = \frac{A}{x} + B x^2$,then $x^2 \frac{d^2 y}{d x^2} =$

Let $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$,$x \in [0, \pi] - \{\frac{\pi}{4}\}$. Then $f(\frac{7\pi}{12}) f''(\frac{7\pi}{12})$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module