Question

Differentiate from first principle:

$\left(iii\right)e^{ax+b}$

Answer 1

Given:

$f\left(x\right)=e^{ax+b}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$


Putting $f\left(x\right)$ in the above expression, we get:


$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{ax+b}{e}^{ah}-e^{ax+b}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{ax+b}(e^{ah}-1)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=a{e}^{ax+b}\underset{h\to 0}{lim}\frac{e^{ah}-1}{ah}$

$\Rightarrow f^{\prime }\left(x\right)=a{e}^{ax+b}\left(1\right)(\because \underset{x\to 0}{lim}\frac{e^x-1}{x}=1)$

$\Rightarrow f^{\prime }\left(x\right)=a{e}^{ax+b}$

Therefore, the derivative of $e^{ax+b}\text{is}a{e}^{ax+b}.$