# How do you differentiate f(x)=cos(3x)*(-2/3sinx) using the product rule?

Mar 27, 2017

$- \frac{2}{3} \cos \left(x\right) \cos \left(3 x\right) + 2 \sin \left(x\right) \sin \left(3 x\right)$

#### Explanation:

Here are some of the rules that need to be used to solve this question:

• Chain rule: $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$, where $f$ and $g$ are functions of $x$
• Product rule: $\frac{d \left(u \cdot v\right)}{\mathrm{dx}} = u \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, where $u$ and $v$ are functions of $x$
• Constant factor rule (actually a special case of the product rule): $\frac{d \left(k u\right)}{\mathrm{dx}} = k \frac{\mathrm{du}}{\mathrm{dx}}$, where $k$ is a constant and u is a function of $x$

First of all, we rewrite the original function to $- \frac{2}{3} \cdot \cos \left(3 x\right) \cdot \sin \left(x\right)$. We will deal with the constant $- \frac{2}{3}$ later and differentiate $\cos \left(3 x\right) \cdot \sin \left(x\right)$ first (permissible by the constant factor rule).

We shall apply the product rule to find the derivative of $\cos \left(3 x\right) \cdot \sin \left(x\right)$. First, we call $u = \cos \left(3 x\right)$ and $v = \sin \left(x\right)$. Then, we need to find $u '$ and $v '$. We know that $v ' = \cos \left(x\right)$. However, to solve $u '$, we need the chain rule.

The chain rule is used for finding composite functions. For the chain rule, we say that $f \left(x\right) = \cos \left(x\right)$ and $g \left(x\right) = 3 x$. The function is then $f \left(g \left(x\right)\right) = \cos \left(3 x\right)$. Applying the chain rule, we find that the derivative of $\cos \left(3 x\right)$ is $\frac{d \left(3 x\right)}{\mathrm{dx}} \cdot \frac{d \cos \left(3 x\right)}{d \left(3 x\right)} = 3 \cdot - \sin \left(3 x\right) = - 3 \sin \left(3 x\right)$.

Now, we have found $u '$. Substituting in the product rule, $\cos \left(3 x\right) \cdot \cos \left(x\right) - \sin \left(x\right) \cdot 3 \sin \left(3 x\right) = \cos \left(x\right) \cos \left(3 x\right) - 3 \sin \left(x\right) \sin \left(3 x\right)$.

Finally, we apply the constant factor rule, i.e. multiplying this function by $- \frac{2}{3}$, to obtain the final answer $- \frac{2}{3} \cos \left(x\right) \cos \left(3 x\right) + 2 \sin \left(x\right) \sin \left(3 x\right)$.