# How do you use the Product Rule to find the derivative of sin(3x) cos(5x) ?

Aug 9, 2015

${y}^{'} = 3 \cos \left(3 x\right) \cdot \cos \left(5 x\right) - 5 \sin \left(3 x\right) \cdot \sin \left(5 x\right)$

#### Explanation:

First, notice that you can write your function as a product of two functions, $f \left(x\right) = \sin \left(3 x\right)$ and $g \left(x\right) = \cos \left(5 x\right)$

$y = f \left(x\right) \cdot g \left(x\right)$

The product rule allows you to differentiate such functions by using the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(\sin \left(3 x\right)\right)\right] \cdot \cos \left(5 x\right) + \sin \left(3 x\right) \cdot \frac{d}{\mathrm{dx}} \left(\cos \left(5 x\right)\right)$

Now, in order to differentiate these two functions, you need to use the chain rule in the form $\sin {u}_{1}$ and $\cos {u}_{2}$, with ${u}_{1} = 3 x$ and ${u}_{2} = 5 x$.

This will get you

$\frac{d}{\mathrm{dx}} \left(\sin {u}_{1}\right) - \left[\frac{d}{{\mathrm{du}}_{1}} \sin {u}_{1}\right] \cdot \frac{d}{\mathrm{dx}} \left({u}_{1}\right)$

$\frac{d}{\mathrm{dx}} \left(\sin {u}_{1}\right) = \cos {u}_{1} \cdot \frac{d}{\mathrm{dx}} \left(3 x\right)$

$\frac{d}{\mathrm{dx}} \left(\sin \left(3 x\right)\right) = \cos \left(3 x\right) \cdot 3$

and

$\frac{d}{\mathrm{dx}} \left(\cos {u}_{2}\right) = \left[\frac{d}{{\mathrm{du}}_{2}} \cos {u}_{2}\right] \cdot \frac{d}{\mathrm{dx}} \left({u}_{2}\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(5 x\right)\right) = - \sin \left(5 x\right) \cdot 5$

Your target derivative will thus be

${y}^{'} = \left[3 \cdot \cos \left(3 x\right)\right] \cdot \cos \left(5 x\right) + \sin \left(3 x\right) \cdot \left[- 5 \sin \left(5 x\right)\right]$

${y}^{'} = \textcolor{g r e e n}{3 \cos \left(3 x\right) \cdot \cos \left(5 x\right) - 5 \sin \left(3 x\right) \cdot \sin \left(5 x\right)}$