How do you differentiate #f(x)=cos3x^(1/3)# using the chain rule?

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Answer 1 · 2016-07-16T21:44:25

#f'(x) = -(sin(3x^(1/3)))/(x^(2/3))#

The chain rule for derivatives states that

#d/dx[f(g(x))] = f'(g(x)) * g'(x)#

In our case, we have

#f(x) = cos(3x^(1/3))#

Thus, by applying the chain rule, we get

#f'(x) = d/dx[cos(3x^(1/3))] = -sin(3x^(1/3)) * 1/cancel(3) * cancel(3)x^(-2/3)#

Simplifying this expression further yields

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

#= -(sin(3x^(1/3)))/(x^(2/3))#