Question

What is the derivative of `sin^2x+cos^2x`?

Answer 1

We have:

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

We can differentiate using the chain rule to get:

`f'(x) = 2sinxd/dx(sinx) + 2cosxd/dx(cosx)`
`" " = 2sinxcosx + 2cosx(-sinx)`
`" " = 2sinxcosx - 2sinxcosx`
`" " = 0 \ \ \ ` QED

However, it really should be clear that the above calculation was completely unnecessary and the result is obvious.

A fundamental trigonometric identity is:

`sin^2A+cos^2A -=1 \ \ \ AA A in RR`

Hence we have:

`f(x) = 1 \ \ \ AA x in RR`

ie, `f(x)` is a constant function, and the derivative of constant is zero.