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