Question

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

Answer 1

We'll need to resort also to chain rule to differentiate the second term `cos(2-z)`

Explanation:

• Product rule: `(ab)'=a'b+ab'`

• Chain rule: `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Renaming `u=2-z` for the second term we can now proceed

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

`(dg(z))/(dz)=z(2cos(2-z)-zsin(2-z))`