Question

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

Answer 1

`d/(dx)(sin^2 x + cos^2 x) = d/(dx)(1) = 0`

Explanation:

The easiest way is to use the Pythagorean identity:

`sin^2x + cos^2x = 1`

Then the derivative of a constant (`1`) is `0`.

Alternatively you can use a combination of the power rule, chain rule and derivatives of `sin x` and `cos x` as follows:

Assume we know:

`d/(dx) sin x = cos x`

`d/(dx) cos x = -sin x`

Power rule:

`d/(dx) x^n = n x^(n-1)`

Chain rule:

`d/(dx) u(v(x)) = u'(v(x)) * v'(x)`

Then:

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