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Differentiation (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Differentiation (p-1)2 Marks
Question
Differentiate the following functions w.r.t. x. : $\frac{e^x}{e^x+1}$
✓

Answer

$
y=\frac{\mathrm{e}^x}{\mathrm{e}^x+1}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{e}^x}{\mathrm{e}^x+1}\right) \text { } \\
&=\frac{\left(\mathrm{e}^x+1\right) \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x\right)-\mathrm{e}^x \frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{e}^x+1\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&=\frac{\left(\mathrm{e}^x+1\right) \mathrm{e}^x-\mathrm{e}^x\left(\mathrm{e}^x+0\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&=\frac{\mathrm{e}^x\left(\mathrm{e}^x+1-\mathrm{e}^x\right)}{\left(\mathrm{e}^x+1\right)^2} \\
&= \frac{\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}
\end{aligned}
$

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