Question

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

Answer 1

`-2/3 cos(x)cos(3x)+2sin(x)sin(3x)`

Explanation:

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

• Chain rule: `(df)/dx=(df)/(dg)*(dg)/dx`, where `f` and `g` are functions of `x`
• Product rule: `(d(u*v))/dx=u*(dv)/(dx)+v*(du)/(dx)`, where `u` and `v` are functions of `x`
• Constant factor rule (actually a special case of the product rule): `(d(ku))/dx=k (du)/dx`, where `k` is a constant and u is a function of `x`

First of all, we rewrite the original function to `-2/3*cos(3x)*sin(x)`. We will deal with the constant `-2/3` later and differentiate `cos(3x)*sin(x)` first (permissible by the constant factor rule).

We shall apply the product rule to find the derivative of `cos(3x)*sin(x)`. First, we call `u=cos(3x)` and `v=sin(x)`. Then, we need to find `u'` and `v'`. We know that `v'=cos(x)`. 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(x)=cos(x)` and `g(x)=3x`. The function is then `f(g(x))=cos(3x)`. Applying the chain rule, we find that the derivative of `cos(3x)` is `(d(3x))/dx*(d cos(3x))/(d(3x))=3*-sin(3x)=-3sin(3x)`.

Now, we have found `u'`. Substituting in the product rule, `cos(3x)*cos(x)-sin(x)*3sin(3x)=cos(x)cos(3x)-3sin(x)sin(3x)`.

Finally, we apply the constant factor rule, i.e. multiplying this function by `-2/3`, to obtain the final answer `-2/3 cos(x)cos(3x)+2sin(x)sin(3x)`.