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

1 Answer
sente
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#

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