Question

25-44 Differentiate.
28. $J(v)=\left(v^{3}-2 v\right)\left(v^{-4}+v^{-2}\right)$

Answer 1

Make things easier by expanding J(v) first:

$J(v) = (v^3-2v) \left(\dfrac1{v^4}+\dfrac1{v^2}\right) = \dfrac1v - \dfrac2{v^3} - \dfrac2v + v = v - \dfrac1v - \dfrac2{v^3}$

Now differentiate term-by-term (power rule):

$J'(v) = \boxed{1 - \dfrac1{v^2} + \dfrac6{v^4}}$

In case you're supposed to use the product rule first, we have

$J'(v) = (3v^2 - 2) \left(\dfrac1{v^4} + \dfrac1{v^2}\right) + (v^3-2v) \left(-\dfrac4{v^5} - \dfrac2{v^3}\right)$

Expanding and simplifying will yield the same result as before.