Question

How do you find the derivative of `sin^2x+sinxcosx`?

Answer 1

`d/dx(sin^2(x)+sin(x)cos(x))=sin(2x)+cos(2x)`

Explanation:

For this problem, we will be using the product rule, the chain rule, and the following properties and derivatives:

• `d/dx(f(x)+g(x)) = f'(x)+g'(x)`

• `d/dx x^2 = 2x`

• `d/dx sin(x) = cos(x)`

• `d/dx cos(x) = -sin(x)`

Proceeding,

`d/dx(sin^2(x)+sin(x)cos(x)) = d/dxsin^2(x) + d/dxsin(x)cos(x)`

`=2sin(x)(d/dxsin(x))+d/dxsin(x)cos(x)`

`=2sin(x)cos(x)+d/dxsin(x)cos(x)`

`=2sin(x)cos(x)+sin(x)(d/dxcos(x))+cos(x)(d/dxsin(x))`

`=2sin(x)cos(x)+sin(x)(-sin(x))+cos(x)(cos(x))`

`=2sin(x)cos(x)+cos^2(x)-sin^2(x)`

`=sin(2x)+cos(2x)`