Question

Compute the derivative of $\tan x.$

Answer 1

Let $f\left(x\right)=\tan x$
We know that $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\tan (x+h)-\tan x}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{h}\left(\tan \right(x+h)-\tan x)$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{h}(\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x})$

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

Using $\left\{\sin \right(A-B)=\sin A\cos B-\cos A\sin B\}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{h}\left(\frac{\sin (x+h-x)}{\cos (x+h)\cos x}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{h}\left(\frac{\left(\sin h\right)}{\cos (x+h)\cos x}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin h}{h}\frac{1}{\cos (x+h)\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin h}{h}\times \underset{h\to 0}{lim}\frac{1}{\cos (x+h)\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=1\times \underset{h\to 0}{lim}\frac{1}{\cos (x+h)\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{\cos (x+h)\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\cos (x+0)\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\cos x\cos x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{{\cos }^{2}x}$
$\Rightarrow f^{\prime }\left(x\right)={\sec }^{2}x$
Hence, $f^{\prime }\left(x\right)={\sec }^{2}x$