How do you differentiate #J(v)= (v^3-2v) (v^-4 + v^-2)#?

1 Answer
Apr 6, 2018

Answer:

#J'(v) = (3v^2-2)(v^(-4)+v^(-2))- (v^3-2v)(4v^(-5) +2v^(-3))#

Explanation:

In order to differentiate #J(v)#, we have to use the #color(red)("Product Rule")#, which states that

#[f(v)*g(v)]' = f'(v)g(v) + f(v)g'(v)#

In our case, #f(v) = v^3 - 2v# and #g(v) = v^(-4) + v^(-2)#

The derivative of both of these functions can be found using the Power rule:

#f(x) = x^k => f'(x) = kx^(k-1)#

#f(v) = v^3-2v#

#:. f'(v) = [v^3-2v]' = [v^3]' - 2[v]' = 3v^2-2#

#g(v) = v^(-4) + v^(-2)#

#:. g'(v) = -4v^(-5) -2v^(-3)#

Therefore, we have:

#J'(v) = f'(v)g(v)+f(v)g'(v)#

#J'(v) = (3v^2-2)(v^(-4)+v^(-2))- (v^3-2v)(4v^(-5) +2v^(-3))#