Differentiate the following functions from first principles : e^ … - Vidyadip

Question

Differentiate the following functions from first principles : $e^{ax+b}.$

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Answer

Let $f(x) = e^{ax+b}\Rightarrow f(x + h) = e^{a(x+h)+b}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}(\text{x}+\text{h})+\text{b}}-\text{e}^{(\text{ax}+\text{b})}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}+\text{b}}\text{e}^{\text{ah}}-\text{e}^{\text{ax}+\text{b}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\text{ax}+\text{b}}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}\times\text{a}$
$=\text{ae}^{\text{ax}+\text{b}} \lim\limits_{\text{h}\rightarrow0}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}$
$=\text{ae}^{\text{ax}+\text{b}}$
So, $\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}+\text{b})=\text{ae}^{\text{ax}+\text{b}}$

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