Question

Find the derivative of the following w.r.t. x at the point indicated against them by using method of first principle :

$\tan \: x \: \: \: \: \: \: \: \: at \: \: \: \: x = \frac{\pi}{4}$


(Please give correct answers with step by step explanation only.)​

Answer 1

hope this answer is helpful and mark as brainliest if u can plzz

Answer 2

The derivative of the following w.r.t. x at the point indicated against them by using method of first principle  is as follows:

Given,

tan x at x = π/4

First principle method of derivation is given by,

$f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}\\ \\\\f'(x) = \lim_{h \to 0} \dfrac{tan(x+h)-tan(x)}{h}\\\\= \lim_{h \to 0} \dfrac{\frac{sin(x+h)}{cos(x+h)}-\frac{sinx}{cosx}}{h}\\\\\\= \lim_{h \to 0} \dfrac{sin(x+h)cosx-sinxcos(x+h)}{hcos(x+h)cosx}\\\\\\= \lim_{h \to 0}\dfrac{\frac{sin(2x+h)+sinh}{2}-\frac{sin(2x+h)-sinh}{2}}{hcos(x+h)cosx}\\\\\\= \lim_{h \to 0} \dfrac{sinh}{hcos(x+h)cosx}\\\\= \lim_{h \to 0} \dfrac{sinh}{h} \times \lim_{h \to 0} \dfrac{1}{cos(x+h)cosx}$

$= 1 \times \dfrac{1}{cosx \times cosx}\\\\= 1 \times \dfrac{1}{cos^2x}\\\\\\=sec^2x$

Hence the derivative of tanx is sec²x

At x = π/4

sec²x = sec²π/4

= sec π/4 × secπ/4

= √2 × √2

= 2