Question

Find the derivative of tan x using first principle of derivatives

Answer 1

From the first principle of derivatives,

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

$=\stackrel{lim}{h\to 0}\frac{tan(x+h)-tanx}{h}$

$=\stackrel{lim}{h\to 0}\frac{\frac{sin(x+h)}{cos(x+h)}-\frac{sinx}{cosx}}{h}$

$=\stackrel{lim}{h\to 0}\frac{cosxsin(x+h)-sinxcos(x+h)}{hcosxcos(x+h)}$

$=\stackrel{lim}{h\to 0}\frac{\frac{sin(2x+h)+sinh}{2}-\frac{sin(2x+h)-sinh}{2}}{hcosxcos(x+h)}$

$=\stackrel{lim}{h\to 0}\frac{sinh}{hcosxcos(x+h)}$

$=\stackrel{lim}{h\to 0}\frac{sinh}{h}\times \stackrel{lim}{h\to 0}\frac{1}{cosxcos(x+h)}$

$=1\times \frac{1}{cosx\times cosx}$

$=\frac{1}{co{s}^{2}x}$

$f^{\prime }\left(x\right)=se{c}^{2}x$