Differentiate the following from first principle:e^ ax+b - Vidyadip

Question

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

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Answer

We have,$\text{f(x)}=\text{e}^{\text{ax}+\text{b}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}\text{(x+h)}+\text{b}}-\text{e}^{\text{ax+b}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}}\times\text{e}^{\text{ah}}\times\text{e}^\text{b}-\text{e}^{\text{ax}}\times\text{e}^\text{b}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{b}\times\text{e}^{\text{ax}}\big(\text{e}^{\text{ax}}-1\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\text{ax+b}}\times\frac{\text{a}\big(\text{e}^{\text{ah}}-1\big)}{\text{a.h}}$
Multiplying Numerator and denominator by a $\bigg[=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{e}^{\text{ah}}-1\big)}{\text{ah}}=1\bigg]$
$\text{ae}^{\text{ax+b}}$

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