How do you find the derivative of #sin(x^3)#?

1 Answer

We use the chain rule.

http://socratic.org/calculus/basic-differentiation-rules/chain-rule

Using the notation provided there, if we define #y(x) = sin(x^3)# and #u=x^3#, we may rewrite #y(x)# as #y(u)=sin(u)#

From the chain rule we know that #dy/dx = (du)/dx (dy)/(du)#. Recall that #u(x) = x^3# and #y(u) = sin(u)#. Therefore, by the power rule, #(du)/dx = 3x^2#, and by the definitions of trigonometric derivatives, #dy/(du) = cos(u)#. Thus:

#dy/dx = (3x^2)(cos (u)) #

Substituting #x^3# back for u yields:

#dy/dx = 3x^2 cos(x^3)#