# What is the derivative of e^3x cos2x?

Aug 6, 2015

${y}^{'} = {e}^{3} \cdot \left[\cos \left(2 x\right) - 2 x \cdot \sin \left(2 x\right)\right]$

#### Explanation:

To differentiate this function, you can use the product rule and the chain rule. Keep in mind that you have

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

So, according to the product rule, you can differentiate a function that takes the form

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

by using the formula

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

In your case, $f \left(x\right) = x$ and $g \left(x\right) = \cos \left(2 x\right)$, which means that you can write

$\frac{d}{\mathrm{dx}} \left(y\right) = {e}^{3} \left(\left[\frac{d}{\mathrm{dx}} \left(x\right)\right] \cdot \cos \left(2 x\right) + x \cdot {\underbrace{\frac{d}{\mathrm{dx}} \left(\cos \left(2 x\right)\right)}}_{\textcolor{red}{\text{use chain rule!}}}\right)$

${y}^{'} = {e}^{3} \cdot \left[1 \cdot \cos \left(2 x\right) + x \cdot \left(- \sin \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)\right)\right]$

${y}^{'} = {e}^{3} \cdot \left[\cos \left(2 x\right) - x \cdot \sin \left(2 x\right) \cdot 2\right]$

${y}^{'} = \textcolor{g r e e n}{{e}^{3} \cdot \left[\cos \left(2 x\right) - 2 x \cdot \sin \left(2 x\right)\right]}$