Question

Differentiate from first principle:
$\left(i\right)sin\sqrt{2x}$

Answer 1

Given:
$f\left(x\right)=sin\sqrt{2x}$

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 in above expression, we get:
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{sin\sqrt{2(x+h)}-sin\sqrt{2x}}{h}$

Applying the formula,

$sinC-sinD=2cos\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right),$

we get



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


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

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

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

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

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

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

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