How do you differentiate #g(z) = z^2cos(z^2-z)# using the product rule?

1 Answer
Dec 24, 2015

Answer:

First, see that the two factors here are #z^2# and #cos(z^2-z)#.
Second, we'll need also the chain rule to derivate the second term.

Explanation:

  • Product rule states that for #y=f(x)g(x)#, #(dy)/(dx)=(dy)/(du)(du)/(dx)#
  • Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Here, we can rename #u=z^2-z# and now apply chain rule when we need to derivate the second term, following the product rule.

So:

#(dg(z))/(dz)=2zcos(z^2-z)+z^2u'sin(u)#

#(dg(z))/(dz)=2zcos(z^2-z)+z^2(2z-1)sin(z^2-z)#

#(dg(z))/(dz)=2z[cos(z^2-z)+(z^2-1)sin(z^2-z)]#