differentiate $y(u)=(u^{-2}+u^{-3})(u^{5}-6u^{2})$.
$y(u)=$

Question

Accepted Answer

# Explanation:
## Step1: Apply product - rule
The product - rule states that if $Y(u)=f(u)g(u)$, then $Y^{\prime}(u)=f^{\prime}(u)g(u)+f(u)g^{\prime}(u)$. Let $f(u)=u^{-2}+u^{-3}$ and $g(u)=u^{5}-6u^{2}$.
## Step2: Differentiate $f(u)$
Using the power - rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $f^{\prime}(u)=-2u^{-3}-3u^{-4}$.
## Step3: Differentiate $g(u)$
Using the power - rule, $g^{\prime}(u)=5u^{4}-12u$.
## Step4: Substitute into product - rule
$$
\begin{align*}
Y^{\prime}(u)&=(-2u^{-3}-3u^{-4})(u^{5}-6u^{2})+(u^{-2}+u^{-3})(5u^{4}-12u)\
&=(-2u^{-3}\times u^{5}-2u^{-3}\times(-6u^{2})-3u^{-4}\times u^{5}-3u^{-4}\times(-6u^{2}))+(u^{-2}\times5u^{4}+u^{-2}\times(-12u)+u^{-3}\times5u^{4}+u^{-3}\times(-12u))\
&=(-2u^{2}+12u^{-1}-3u + 18u^{-2})+(5u^{2}-12u^{-1}+5u - 12u^{-2})\
&=(-2u^{2}+5u^{2})+(12u^{-1}-12u^{-1})+(-3u + 5u)+(18u^{-2}-12u^{-2})\
&=3u^{2}+2u + 6u^{-2}
\end{align*}
$$
# Answer:
$3u^{2}+2u+\frac{6}{u^{2}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply product - rule

Step2: Differentiate $f(u)$

Step3: Differentiate $g(u)$

Step4: Substitute into product - rule

Answer: