Question

How do you find the derivative of `y=5/(2x)^2+2cosx`?

Answer 1

`dy/dx=5/(2x^3)-2sinx`

Explanation:

Once we rewrite, we will need the rules:

• `d/dxcf(x)=cd/dxf(x)`
• `d/dxx^n=nx^(n-1)`
• `d/dxcosx=-sinx`

So first we see that

`y=5/(4x^2)+2cosx`

`y=5/4x^-2+2cosx`

So then

`dy/dx=5/4(-2)x^-3+2(-sinx)`

`dy/dx=5/(2x^3)-2sinx`