Question

Differentiate from first principle:
$\left(ii\right)cos\sqrt{x}$

Answer 1

Given:
$f\left(x\right)=cos\sqrt{x}$

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;

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

Applying the formula,

$cosC-cosD=-2sin\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right)$

we get

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-2sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=-2\underset{h\to 0}{lim}\frac{sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}\times \frac{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}\times sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=-\underset{h\to 0}{lim}\frac{sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}\times \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\times sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=-\underset{h\to 0}{lim}\frac{sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}\times \frac{1}{(\sqrt{x+h}+\sqrt{x})}\times sin\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=-\left(1\right)\times \frac{1}{(\sqrt{x+0}+\sqrt{x})}\times sin\left(\frac{\sqrt{x+0}+\sqrt{x}}{2}\right)$

$[\because \underset{h\to 0}{lim}\frac{sin\left(h\right)}{h}=1]$

$\Rightarrow f^{\prime }\left(x\right)=-\frac{sin\left(\sqrt{x}\right)}{2\sqrt{x}}$

Therefore, the derivative of $cos\sqrt{x}$ is $-\frac{sin\sqrt{x}}{2\sqrt{x}}$