differentiate the function.
$f(y)=(\frac{1}{y^{2}}-\frac{9}{y^{4}})(y + 3y^{3})$
$f(y)=$

Question

Accepted Answer

# Explanation:
## Step1: Rewrite the function
First, rewrite $F(y)$ as $F(y)=(y^{- 2}-9y^{-4})(y + 3y^{3})$.
## Step2: Apply the product - rule
The product - rule states that if $F(y)=u(y)v(y)$, then $F'(y)=u'(y)v(y)+u(y)v'(y)$. Let $u(y)=y^{-2}-9y^{-4}$ and $v(y)=y + 3y^{3}$. Then $u'(y)=-2y^{-3}+36y^{-5}$ and $v'(y)=1 + 9y^{2}$.
## Step3: Calculate $F'(y)$
$$
\begin{align*}
F'(y)&=(-2y^{-3}+36y^{-5})(y + 3y^{3})+(y^{-2}-9y^{-4})(1 + 9y^{2})\
&=(-2y^{-3}\times y-2y^{-3}\times3y^{3}+36y^{-5}\times y+36y^{-5}\times3y^{3})+(y^{-2}\times1+y^{-2}\times9y^{2}-9y^{-4}\times1-9y^{-4}\times9y^{2})\
&=(-2y^{-2}-6 + 36y^{-4}+108y^{-2})+(y^{-2}+9-9y^{-4}-81y^{-2})\
&=(-2y^{-2}+108y^{-2}+y^{-2}-81y^{-2})+( - 6 + 9)+(36y^{-4}-9y^{-4})\
&=( - 2 + 108+1 - 81)y^{-2}+3+(36 - 9)y^{-4}\
&=26y^{-2}+3 + 27y^{-4}\
&=\frac{26}{y^{2}}+3+\frac{27}{y^{4}}
\end{align*}
$$
# Answer:
$\frac{26}{y^{2}}+3+\frac{27}{y^{4}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rewrite the function

Step2: Apply the product - rule

Step3: Calculate $F'(y)$

Answer: