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Question: If y = x + e<sup>x</sup>, then the value of \(\frac{d^{2}x}{dy^{2}}\) at x = 1, will be-...

If y = x + ex, then the value of d2xdy2\frac{d^{2}x}{dy^{2}} at x = 1, will be-

A

e

B

−e(1+e)3\frac{- e}{(1 + e)^{3}}

C

−e(1+e)\frac{- e}{(1 + e)}

D

−e(1+e)2\frac{- e}{(1 + e)^{2}}

Answer

−e(1+e)3\frac{- e}{(1 + e)^{3}}

Explanation

Solution

dydx\frac{dy}{dx} = 1 + ex

̃ dxdy\frac{dx}{dy} = 11+ex\frac{1}{1 + e^{x}}

d2xdy2\frac{d^{2}x}{dy^{2}} = ddx(11+ex).dxdy\frac{d}{dx}\left( \frac{1}{1 + e^{x}} \right).\frac{dx}{dy} = −ex(1+ex)2.1(1+ex)\frac{- e^{x}}{(1 + e^{x})^{2}}.\frac{1}{(1 + e^{x})}

\  d2xdy2∣x=1\left. \ \frac{d^{2}x}{dy^{2}} \right|_{x = 1} = −e(1+e)3\frac{- e}{(1 + e)^{3}}