How do you calculate the derivative of the function #f(x)=cos(x^3+x^2+1)#?

1 Answer
Jim H
Mar 6, 2015

Use the chain rule.

#d/dx(f(g(x))=f'(g(x))g'(x)#

In this problem, the outermost function (#f#) is the cosine function.

The derivative of the cosine function is the opposite of the sign function.

So we take #f'# (the opposite of the sine) evaluated at the inside function (#g(x)# which is #x^3+x^2+1#) TIMES the derivative of the inside function.

The derivative is #-sin(x^3+x^2+1)# times #(3x^2+2x)# (the derivative of the inside function).

#f'(x)=(-sin(x^3+x^2+1))*(3x^2+2x)=-(3x^2+2x)sin(x^3+x^2+1)#.

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