If y = x + e^x , then at x = 1 , d^2 x d y^2 is equal to - Vidyadip

MCQ

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

A
$e$
✓
$\frac{{ - e}}{{{{\left( {1 + e} \right)}^{^3}}}}$
C
$\frac{{ - e}}{{\left( {1 + e} \right)}}$
D
$\frac{{ - e}}{{{{\left( {1 + e} \right)}^2}}}$

✓

Answer

Correct option: B.

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

b
$\frac{d y}{d x}=1+e^{x}$

$\Rightarrow \frac{d x}{d y}=\frac{1}{1+e^{x}}$

$\frac{{{d^2}x}}{{d{y^2}}} = \frac{d}{{dx}}\left( {\frac{1}{{1 + {e^x}}}} \right)\frac{{dx}}{{dy}}$

$ = \frac{{ - {e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cdot \frac{1}{{\left( {1 + {e^x}} \right)}}$

${\left. {\therefore \frac{{{d^2}x}}{{d{y^2}}}} \right|_{x = 1}} = - \frac{e}{{{{(1 + e)}^3}}}$

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