Question

What is the derivative of `f(x) = sin^2(x)+cos^2(x)`?

Answer 1

`d/dx sin^2(x) + cos^2(x) = 0`

Explanation:

There are multiple ways of going about this. The easiest is to notice that `sin^2(x) + cos^2(x) = 1` and so
`d/dx sin^2(x) + cos^2(x) = d/dx 1 = 0`

However, without this, we can solve this using the chain rule:
`d/dxf(g(x)) = f'(g(x))g'(x)`
together with the derivatives
`d/dxsin(x) = cos(x)`
`d/dxcos(x) = -sin(x)`
`d/dxx^n = nx^(n-1)`

Proceeding in this manner, we apply the chain rule to get
`d/dx sin^2(x) = 2sin(x)(d/dxsin(x)) = 2sin(x)cos(x)`
and
`d/dxcos^2(x) = 2cos(x)(d/dxsin(x)) = -2sin(x)cos(x)`

Thus
`d/dx sin^2(x) + cos^2(x) = 2sin(x)cos(x) - 2sin(x)cos(x) = 0`