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

1 Answer
Nov 14, 2015

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