Question

How do you find the derivative of cos2x/x ?

Answer 1

`(-2xsin(2x)-cos(2x))/x^2`

Explanation:

We'll be able to find the derivative with respect to `x` of`cos(2x)/x` by an application of the Chain Rule and the Quotient Rule.

Since we have a fraction essentially, it will help to use the Quotient Rule, below:

`(f'(x)*g(x)-f(x)*g'(x))/g(x)^2`

Let's define our functions:
`f(x)=cos(2x)`
`g(x)=x`

If the Quotient or Chain Rules seem foreign to you, I encourage you to google them or get some practice at them on Khan Academy.

To find the derivative of f`(x)`, since it is a composite function, we have to use to Chain Rule, stated below:

`f'(h(x))*h'(x)`

The Chain Rule is essentially taking the derivative of the outside function with respect to the inside, and multiplying it by the derivative of the inner function. In `cos(2x)`, `cosx` is the outside function (`f(x)`), and `2x` is the inner function (`(h(x)`).

`f'(h(x))*h'(x)= -sin(2x)*2= color(blue)(-2sin(2x))`

Now, we can return to the Quotient Rule. What I have in blue is `f'(x)` and the derivative `g(x)`, which is `x`, will be `1`. Plugging into the Quotient Rule, we get:

`(-2sin(2x)*x-cos(2x)*1)/x^2`

Which can be simplified to:

`(-2xsin(2x)-cos(2x))/x^2`

Again, I strongly encourage you to Google the Quotient and Chain Rules or comment below if anything seems confusing. Hope this helps!