Question

part 1 of 4
to find the derivative of the product $f(y) = \left(\frac{1}{y^2} - \frac{4}{y^4}\
ight)(y + 6y^3)$, we will us
$\frac{d}{dy}f(y)g(y) = f(y)g(y) + f(y)g(y)$.
first, we rewrite the first term of the product as follows.
$\frac{1}{y^2} - \frac{4}{y^4} = y^{-2} - 4y^{-4}$

part 2 of 4
the derivative of the first term, $\frac{1}{y^2} - \frac{4}{y^4} = y^{-2} - 4y^{-4}$, is
$-2y^{-3} + 16y^{-5}$

part 3 of 4
next, the derivative of the second term, $y + 6y^3$, is
blank box

Explanation:

Step1: Apply power rule to each term

Derivative of $y$ is $1$, derivative of $6y^3$ is $6*3y^{2}=18y^2$.

Step2: Sum the derivatives

Add the results from Step1.

Answer:

$1 + 18y^2$