Question

Using first principle, find the derivative of $tan\sqrt{x}.$

Answer 1

Let $f\left(x\right)=tan\sqrt{x}$

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

$=\underset{h\to 0}{lim}\frac{tan\sqrt{x+h}-tan\sqrt{x}}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{sin\sqrt{x+h}}{cos\sqrt{x+h}}-\frac{sin\sqrt{x}}{cos\sqrt{x}}}{h}$

$=\underset{h\to 0}{lim}\frac{sin\sqrt{x+h}cos\sqrt{x}-sin\sqrt{x}cos\sqrt{x+h}}{hcos\left(\sqrt{x+h}\right)cos\sqrt{x}}$

$=\underset{h\to 0}{lim}\frac{sin\left[\right(\sqrt{x+h})-\sqrt{x}]}{hcos\sqrt{x}cos(\sqrt{(}x+h))}[\because sinAcosB-cosAsinB=\sin (A-B\left)\right]$

$=\underset{h\to 0}{lim}\frac{sin(\sqrt{x+h}-\sqrt{x})}{hcos\sqrt{x}cos\left(\sqrt{x+h}\right)}\times \frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}-\sqrt{x}}$

[multipying numerator and denominator by $\sqrt{x+h}-\sqrt{x}]$

$=\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}x}{hcos\sqrt{x}cos\left(\sqrt{x+h}\right)}\times \underset{h\to 0}{lim}\left(\frac{sin(\sqrt{x+h}-\sqrt{x})}{\sqrt{x+h}-\sqrt{x}}\right)$

$=\frac{1}{co{s}^2\sqrt{x}}\times \underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times 1[\because \underset{h\to 0}{lim}\frac{sinh}{h}=1]$

$=se{c}^2\sqrt{x}\times \underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=se{c}^2\sqrt{x}\times \underset{h\to 0}{lim}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$

$=se{c}^2\sqrt{x}\times \underset{h\to 0}{lim}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$

$=se{c}^2\sqrt{x}\times \frac{1}{2\sqrt{x}}=\frac{se{c}^2\sqrt{x}}{2\sqrt{x}}$