# What is the derivative of y=tan(x^2) ?

Aug 29, 2014

The derivative of $y = \tan \left({x}^{2}\right)$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x\right) \left[{\sec}^{2} \left({x}^{2}\right)\right]$

In solving this problem, we will have to make use of the chain rule, which states that for a function $y = f \left(g \left(x\right)\right)$ (that is, $y$ is a function which is itself a function of another function), $y ' \left(x\right) = \left(g ' \left(x\right)\right) \cdot f ' \left(g \left(x\right)\right)$.

In this case, $f \left(g \left(x\right)\right) = \tan \left({x}^{2}\right)$. Our $g \left(x\right) = {x}^{2}$ and our $f \left(g\right) = \tan \left(g \left(x\right)\right)$. Thus, if our function $f \left(g\right)$ were $f \left(x\right)$ we would have $f \left(x\right) = \tan \left(x\right)$.

Via the power rule we know that $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 x$, and via our definitions of the derivatives of trigonometric functions, we know that $f ' \left(x\right) = {\sec}^{2} \left(x\right)$,or $f ' \left(g\right) = {\sec}^{2} \left(g\right)$

Thus, by using the chain rule shown above, we obtain:

$y ' \left(x\right) = \left(2 x\right) {\sec}^{2} \left({x}^{2}\right)$