Differentiate the following from the first principlee^ x^2+1 - Vidyadip

Question

Differentiate the following from the first principle$\text{e}^{\text{x}^2+1} $

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Answer

We have, $\text{f}(\text{x})=\text{e}^{\text{x}^2+1}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{(\text{x}+\text{h})^2}+1-\text{e}^ {\text{x}^2+1}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+\text{h}^2+\text{2xh}+1}-\text{e}^ {\text{x}^2+1}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+1}(\text{e}^{\text{2xh}}.\text{e}^{\text{h}^2}-1)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+1}(\text{e}^{\text{2xh}+\text{h}^2}-1)}{\text{2xh}+\text{h}^2}\times\frac{\text{2xh}+\text{h}^2}{\text{h}}$
$\because\text{h}\rightarrow0$
$\Rightarrow\text{2xh}+\text{h}^2=0$
and $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\text{x}^2+1}.1\times\text{2x}+\text{h}$
$=\text{2xe}^{\text{x}^2+1}$

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