Question

How do you find the derivative of $y=\tan \left(x\right)$ using first principle ?

Answer 1

Solve by using first principle:

The first principle of calculus can be given as,

$f\text{'}\left(x\right)=\underset{x\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

We have,

$f\left(x\right)=y=\tan \left(x\right)$

Thus,

$\begin{array}{cc}f\text{'}\left(x\right)=\underset{x\to 0}{\lim }\frac{\tan \left(x+h\right)-\tan \left(x\right)}{h}& \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\sin \left(x+h\right)}{\cos \left(x+h\right)}}-{\displaystyle \frac{\sin \left(x\right)}{\cos \left(x\right)}}}{h}& [\because \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}]\\ =\underset{x\to 0}{\lim }\frac{\sin \left(x+h\right)·\cos \left(x\right)-\cos \left(x+h\right)\sin \left(x\right)}{h\cos \left(x+h\right)\cos \left(x\right)}& \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\sin \left(x+h+x\right)+\sin \left(x+h-x\right)}{2}}-\left[{\displaystyle \frac{\sin \left(x+h+x\right)-\sin \left(x+h-x\right)}{2}}\right]}{h\cos \left(x+h\right)\cos \left(x\right)}& [\because \sin \left(a\right)\cos \left(b\right)=\frac{\sin \left(a+b\right)+\sin \left(a-b\right)}{2},\cos \left(a\right)\sin \left(b\right)=\frac{\sin \left(a+b\right)-\sin \left(a-b\right)}{2}]\\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\sin \left(2x+h\right)+\sin \left(h\right)}{2}}-{\displaystyle \frac{\sin \left(2x+h\right)-\sin \left(h\right)}{2}}}{h\cos \left(x+h\right)\cos \left(x\right)}& \\ =\underset{x\to 0}{\lim }\frac{\sin \left(h\right)}{h\cos \left(x+h\right)\cos \left(x\right)}& \\ =\underset{x\to 0}{\lim }\frac{\sin \left(h\right)}{h}·\frac{1}{\cos \left(x+h\right)\cos \left(x\right)}& \\ =\underset{x\to 0}{\lim }\frac{1}{\cos \left(x+h\right)\cos \left(x\right)}& [\because \underset{x\to 0}{\lim }\frac{\sin \left(x\right)}{x}=1]\\ =\frac{1}{\cos \left(x\right)\cos \left(x\right)}& \\ ={\sec }^2\left(x\right)& \end{array}$

Therefore, derivative of $\tan \left(x\right)$ is ${\sec }^2\left(x\right)$.