# How do you find the derivative of y=tan^2(x) ?

##### 1 Answer
Aug 29, 2014

The derivative of $y = {\tan}^{2} \left(x\right)$ is $y ' \left(x\right) = 2 {\sec}^{2} \left(x\right) \tan \left(x\right)$

To find the derivative, we will need to make use of two properties. The first is the Product Rule, which states that given a function $f \left(x\right)$ that is itself the product of other functions $g \left(x\right)$ and $h \left(x\right)$, that is, $f \left(x\right) = g \left(x\right) h \left(x\right)$, the derivative $f ' \left(x\right)$ equals $g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$. In other words, the derivative of a function that is the product of two other functions is equal to the sum of the two expressions formed by the product of each function with the other function's derivative.

Our second property consists of the definitions of the derivatives of the six basic trigonometric functions. Specifically, we only require the derivative of $\tan \left(x\right)$, which is $\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$. This will be accepted without proof, but a proof does in fact exist.

For this calculation, we will represent $y = {\tan}^{2} \left(x\right)$ with its equivalent, $y = \tan \left(x\right) \tan \left(x\right)$. This will allow us to use the product rule. We declare $f \left(x\right) = y \left(x\right) = g \left(x\right) h \left(x\right) = \tan \left(x\right) \tan \left(x\right)$, and by utilizing $\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$ along with $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$, we obtain...

$f ' \left(x\right) = {\sec}^{2} \left(x\right) \tan \left(x\right) + \tan \left(x\right) {\sec}^{2} \left(x\right) = 2 \tan \left(x\right) {\sec}^{2} \left(x\right)$