Differentiate the following functions from first principles: eax+… - Vidyadip

Question

Differentiate the following functions from first principles:

e^ax+b.

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Answer

Let f(x) = e^ax+b

⇒ 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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