MCQ
If $y = x + e^x$ then $\frac{{{d^2}x}}{{d{y^2}}}$ is :
- A$e^x$
- ✓$ - \frac{{{e^x}}}{{{{\left( {1\, + \,{e^x}} \right)}^3}}}$
- C$-\frac{{{e^x}}}{{{{\left( {1\, + \,{e^x}} \right)}^2}}}$
- D$\frac{{ - 1}}{{{{\left( {1\, + \,{e^x}} \right)}^3}}}$
✓
Answer
Correct option: B.
$ - \frac{{{e^x}}}{{{{\left( {1\, + \,{e^x}} \right)}^3}}}$
b
$y=x+e^{x}$
$y=x+e^{x}$
$d y / d x=1+e^{x}$
$\Rightarrow \quad \frac{d x}{d y}=\frac{1}{1+e^{x}}$
$\Rightarrow \quad \frac{d^{2} x}{d y^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{2}} \cdot \frac{d x}{d y}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$
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