# How do you differentiate x^1.7cosx?

Jul 30, 2018

$\frac{d}{\mathrm{dx}} \left({x}^{1.7} \cos \left(x\right)\right) = 1.7 {x}^{0.7} \cos \left(x\right) - {x}^{1.7} \sin \left(x\right)$

#### Explanation:

This problem is an example of the Product Rule where

$f \left(x\right) = {x}^{1.7}$ and $g \left(x\right) = \cos \left(x\right)$

Their respective derivatives are

$f ' \left(x\right) = 1.7 {x}^{0.7}$ and $g ' \left(x\right) = - \sin \left(x\right)$

The Product Rule states that

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f \left(x\right) \cdot g ' \left(x\right) + f ' \left(x\right) \cdot g \left(x\right)$

Plugging the derivatives and original functions in gives

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right)\right) = {x}^{1.7} \left(- \sin \left(x\right)\right) + 1.7 {x}^{0.7} \cos \left(x\right)$

Rewriting for simplification and ease of reading gives

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right)\right) = 1.7 {x}^{0.7} \cos \left(x\right) - {x}^{1.7} \sin \left(x\right)$