Question

derivative of tan²x by first principle​

Answer 1

Given:

tan²x

To find:

We have to find the derivative of tan²x by first principle method

Solution:

From the first principle method $f^{'}(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

here f(x)= tan²x

Now proceeding further,

$f^{'}(x)= \lim_{h \to 0} \frac{tan^{2} (x+h)-tan^{2} x}{h}$

$f^{'}(x)= \lim_{h \to 0} \frac{[tan(x+h)+tanx][tan(x+h)-tanx]}{h}$

since we have the formula, $tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$

$f^{'}(x)= \lim_{h \to 0} \frac{[tan(x+h)+tanx][tan(x+h-x )(1+tan(x+h)tanx)]}{h}$

$f^{'}(x)= \lim_{h \to 0} \frac{[tan(x+h)+tanx][1+tan(x+h)tanx]tanh}{h}$

We have the formula,$\lim_{h \to 0} \frac{tanx}{x} =1$ and h=0 so we get,

$f^{'}(x)= [tan(x+0)+tanx][1+tan(x+0)tanx]$

$f^{'}(x)= [2tanx][1+tan^{2} x]$         ( ∵ $sec^{2}x-tan^{2}x =1$ )

$f^{'}(x)= [2tanx][sec^{2} x]$

$f^{'}(x)=2.tanx.sec^{2}x$

∴Derivative of tan²x by method of first principle is  2tanx.sec²x