# What is the Product Rule for derivatives?

##### 1 Answer
Aug 29, 2014

The product rule for derivatives states that given a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, the derivative of the function is $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

The product rule is used primarily when the function for which one desires the derivative is blatantly the product of two functions, or when the function would be more easily differentiated if looked at as the product of two functions. For example, when looking at the function $f \left(x\right) = {\tan}^{2} \left(x\right)$, it is easier to express the function as a product, in this case namely $f \left(x\right) = \tan \left(x\right) \tan \left(x\right)$.

In this case, expressing the function as a product is easier because the basic derivatives for the six primary trig functions ($\sin \left(x\right) , \cos \left(x\right) , \tan \left(x\right) , \csc \left(x\right) , \sec \left(x\right) , \cot \left(x\right)$) are known, and are, respectively, $\cos \left(x\right) , - \sin \left(x\right) , {\sec}^{2} \left(x\right) , - \csc \left(x\right) \cot \left(x\right) , \sec \left(x\right) \tan \left(x\right) , - {\csc}^{2} \left(x\right)$

However, the derivative for $f \left(x\right) = {\tan}^{2} \left(x\right)$ is not one of the elementary 6 trigonometric derivatives. Thus, we consider $f \left(x\right) = {\tan}^{2} \left(x\right) = \tan \left(x\right) \tan \left(x\right)$ so that we can deal with $\tan \left(x\right)$, for which we know the derivative. Utilizing the derivative of $\tan \left(x\right)$, namely $\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$, and the Chain Rule $\frac{\mathrm{df}}{\mathrm{dx}} = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$, we obtain:

$f ' \left(x\right) = \left[\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right)\right] \tan \left(x\right) + \tan \left(x\right) \left[\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right)\right]$

$\frac{d}{\mathrm{dx}} \tan \left(x\right) = {\sec}^{2} \left(x\right)$, so...

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