What is the derivative of #cos(sin(x))#?

1 Answer
sente
Jul 7, 2016

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

Explanation:

The chain rule states that given two functions #f# and #g#, we have #(f@g)'(x) = f'(g(x))*g'(x)#.

In this case, letting #f(x) = cos(x)# and #g(x) = sin(x)#, we have

#(f@g)(x) = cos(sin(x))#
#f'(x) = -sin(x)#
#g'(x) = cos(x)#

Thus

#d/dxcos(sin(x)) = (f@g)'(x)#

#= f'(g(x))g'(x)#

#=-sin(sin(x))cos(x)#

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