Question

part 3 of 4
next, the derivative of the second term, ( y + 6y^3 ), is
( 1 + 18y^2 ) (\boldsymbol{checkmark}) ( 1 + 18y^2 ).

part 4 of 4
now, using the product rule,
( f(y) = left( square + 16y^{-5}
ight)(y + 6y^3) + (y^{-2} - 4y^{-4})left( 1 + square
ight) ).

and after expanding the parentheses and simplifying, we conclude that the derivative is as follows.
( f(y) = square )

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Explanation:

Step1: Find derivative of first term

First term is $y^{-2} - 4y^{-4}$. Derivative: $\frac{d}{dy}(y^{-2}) - \frac{d}{dy}(4y^{-4}) = -2y^{-3} + 16y^{-5}$. So first blank is $-2y^{-3}$.

Step2: Identify second blank

From Part3, derivative of $y + 6y^3$ is $1 + 18y^2$. So second blank is $18y^2$.

Step3: Expand and simplify

Substitute blanks into $F'(y)$:
$F'(y) = (-2y^{-3} + 16y^{-5})(y + 6y^3) + (y^{-2} - 4y^{-4})(1 + 18y^2)$
Expand first product: $-2y^{-2} -12y^{0} +16y^{-4} +96y^{-2} = 94y^{-2} -12 +16y^{-4}$
Expand second product: $y^{-2} +18y^{0} -4y^{-4} -72y^{-2} = -71y^{-2} +18 -4y^{-4}$
Add both results: $(94y^{-2}-71y^{-2}) + (-12+18) + (16y^{-4}-4y^{-4}) = 23y^{-2} +6 +12y^{-4}$

Answer:

First blank: $-2y^{-3}$
Second blank: $18y^2$
Simplified derivative: $23y^{-2} + 12y^{-4} + 6$ (or $\frac{23}{y^2} + \frac{12}{y^4} + 6$)