Question

Differentiate from first principle:

$\left(xi\right)a^{\sqrt{x}}$

Answer 1

Given:

$f\left(x\right)=a^{\sqrt{x}}$

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 above expression, we get:

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

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{a^{\sqrt{x}}(a^{\sqrt{(x+h)}-\sqrt{x}}-1)}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{a^{\sqrt{x}}(a^{\left(\frac{(x+h)-x}{\sqrt{(x+h)}+\sqrt{x}}\right)}-1)}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{a^{\sqrt{x}}(a^{\left(\frac{h}{\sqrt{(x+h)}+\sqrt{x}}\right)}-1)}{\frac{h}{\sqrt{(x+h)}+\sqrt{x}}}\times \frac{1}{\sqrt{(x+h)}+\sqrt{x}}$

$\Rightarrow f^{\prime }\left(x\right)=a^{\sqrt{x}}{\log }_{e}a\times \frac{1}{\sqrt{(x+0)}+\sqrt{x}}$

$[\because \underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a]$

$\Rightarrow f^{\prime }\left(x\right)=\frac{a^{\sqrt{x}}}{2\sqrt{x}}{\log }_{e}a$

Therefore, the derivative of $a^{\sqrt{x}}$ is $\frac{a^{\sqrt{x}}}{2\sqrt{x}}{\log }_{e}a$