Question

differentiation of tan Square x​

Answer 1

$\textbf{Formula used:}$

$\left\begin{array}{|c|c|}\cline{1-2}1.&\frac{d(x^n)}{dx}=nx^{n-1}\\\cline{1-2}2.&\frac{d(tanx)}{dx}=sec^2x\\\cline{1-2}\end{array}\right$

$\text{I have applied chain rule to differentiate the given function}$

$\textbf{Chain rule:}$

$\boxed{\bf\frac{d(f[g(x)])}{dx}=f'[g(x)]\;g'(x)}$

$\text{Let }y=tan^2x$

$\implies\,y=(tanx)^2$

$\text{Now,}$

$\displaystyle\frac{dy}{dx}=\frac{d((tanx)^2)}{dx}$

$\displaystyle\frac{dy}{dx}=2\,tanx\,\frac{d(tanx)}{dx}$

$\displaystyle\frac{dy}{dx}=2\,tanx\,sec^2x$

$\therefore\textbf{The derivative of \bf\,tan^2x is \bf\,2\,tanx\,sec^2x }$

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