How do you differentiate #f(z) = z^3sin^2(2z)# using the product rule?

1 Answer
Mar 17, 2016

Answer:

Derivative of #f(z)# is #z^2sin(2z){4zcos(2z)+3sin(2z)}#

Explanation:

Product rule states that derivative of a function #f(x)# which is a product of two functions #g(x)# and #h(x)# i.e. #f(x)=g(x)*h(x)# is given by

#d/(dx)f(x)=g(x)*d/(dx)h(x)+d/(dx)g(x)*h(x)#

We will also need here the concept of derivative of a function of a function, say #d/(dx)f(g(x))=(df)/(dg)*(dg)/(dx)#

Hence, derivative of #f(z)=z^3sin^2(2z)# is given by

#d/(dx)f(z)=z^3*d/(dx)sin^2(2z)+d/(dx)z^3*sin^2(2z)#

= #z^3*2sin(2z)*cos(2z)*2+3z^2*sin^2(2z)#

= #4z^3sin(2z)cos(2z)+3z^2sin^2(2z)#

= #z^2sin(2z){4zcos(2z)+3sin(2z)}#