Question

to find the derivative of the product $f(y) = \left(\frac{1}{y^2} - \frac{4}{y^4}\
ight)(y + 6y^3)$, we will use the product rule: $\frac{d}{dy}f(y)g(y) = f(y)g(y) + f(y)g(y)$. 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 $-2y^{\square} + \square y^{\square}$.

Explanation:

Step1: Differentiate $y^{-2}$

Use power rule: $\frac{d}{dy}(y^{-2}) = -2y^{-3}$. The exponent is $-3$.

Step2: Differentiate $-4y^{-4}$

Use power rule: $\frac{d}{dy}(-4y^{-4}) = (-4)(-4)y^{-5} = 16y^{-5}$. Coefficient is $16$, exponent is $-5$.

Answer:

$-2y^{-3} + 16y^{-5}$
(Blanks filled: first blank $-3$, second blank $16$, third blank $-5$)