Differentiate the following from the first principlee^ ax+b

Question

Differentiate the following from the first principle$\text{e}^{\sqrt{\text{ax}+\text{b}}}$

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Answer

We have, $\text{f}(\text{x})=\text{e}^{\sqrt{\text{ax}+\text{b}}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}}-\text{e}^{\sqrt{\text{ax}+\text{b}}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h}+\text{b})}}\big(\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}-1\big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h}+\text{b})}}\big(\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}-1\big)}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}{\text{h}}$
Multiplying numerator and denominator by $\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
$\because\text{h}\rightarrow0\Rightarrow\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
and $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$
$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{ax}+\text{b}}}\times1\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}+\sqrt{\text{ax}+\text{b}}}{\text{h}}$
Again Multiplying numerator and denominator by $\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{ax}+\text{b}}}\times\frac{{\text{a}(\text{x}+\text{h})+\text{b}}-{\text{ax}+\text{b}}}{\text{h}}\times\frac{1}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}+\sqrt{\text{ax}+\text{b}}}$
$=\frac{\text{e}^{\sqrt{\text{ax}+\text{b}}}\times\text{a}}{2\sqrt{\text{ax}+\text{b}}}$
$=\frac{\text{ae}^{\sqrt{\text{ax}+\text{b}}}}{2\sqrt{\text{ax}+\text{b}}}$