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

##### 1 Answer

May 16, 2017

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,