MCQ
જો $y = x + e^x$ તો $x = 1$ આગળ $\frac{{{d^2}x}}{{d{y^2}}}$ મેળવો.
- A$e$
- B$\frac{{ - e}}{{{{\left( {1 + e} \right)}^{^3}}}}$
- C$\frac{{ - e}}{{\left( {1 + e} \right)}}$
- D$\frac{{ - e}}{{{{\left( {1 + e} \right)}^2}}}$
$\Rightarrow \frac{d x}{d y}=\frac{1}{1+e^{x}}$
$\frac{{{d^2}x}}{{d{y^2}}} = \frac{d}{{dx}}\left( {\frac{1}{{1 + {e^x}}}} \right)\frac{{dx}}{{dy}}$
$ = \frac{{ - {e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cdot \frac{1}{{\left( {1 + {e^x}} \right)}}$
${\left. {\therefore \frac{{{d^2}x}}{{d{y^2}}}} \right|_{x = 1}} = - \frac{e}{{{{(1 + e)}^3}}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.