Question

1. -/1 points differentiate. $y = e^{p}(p + psqrt{p})$ $y=$ resources read it submit answer 8. -/1 points differentiate. $f(t)=\frac{sqrt3{t}}{t - 9}$ $f(t)=$

Explanation:

Step1: Simplify the first - function

First, simplify $y = e^{p}(p + p\sqrt{p})$. Since $p\sqrt{p}=p\cdot p^{\frac{1}{2}}=p^{\frac{3}{2}}$, then $y = e^{p}(p + p^{\frac{3}{2}})=e^{p}p+e^{p}p^{\frac{3}{2}}$.

Step2: Apply the product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $p$, then $y'=u'v + uv'$.
For $y_1 = e^{p}p$, let $u = e^{p}$ and $v = p$. Then $u'=e^{p}$ and $v' = 1$. So $y_1'=e^{p}\cdot p+e^{p}\cdot1=e^{p}(p + 1)$.
For $y_2 = e^{p}p^{\frac{3}{2}}$, let $u = e^{p}$ and $v = p^{\frac{3}{2}}$. Then $u'=e^{p}$ and $v'=\frac{3}{2}p^{\frac{1}{2}}$. So $y_2'=e^{p}p^{\frac{3}{2}}+e^{p}\cdot\frac{3}{2}p^{\frac{1}{2}}=e^{p}p^{\frac{1}{2}}(p+\frac{3}{2})$.
Then $y'=y_1'+y_2'=e^{p}(p + 1)+e^{p}p^{\frac{1}{2}}(p+\frac{3}{2})=e^{p}(p + 1+\sqrt{p}(p+\frac{3}{2}))$.

Step3: Differentiate the second - function

For $f(t)=\frac{\sqrt[3]{t}}{t - 9}=\frac{t^{\frac{1}{3}}}{t - 9}$, use the quotient - rule. The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$.
Let $u = t^{\frac{1}{3}}$, then $u'=\frac{1}{3}t^{-\frac{2}{3}}$, and $v=t - 9$, then $v' = 1$.
So $f'(t)=\frac{\frac{1}{3}t^{-\frac{2}{3}}(t - 9)-t^{\frac{1}{3}}\cdot1}{(t - 9)^{2}}=\frac{\frac{t - 9}{3t^{\frac{2}{3}}}-t^{\frac{1}{3}}}{(t - 9)^{2}}=\frac{\frac{t - 9-3t}{3t^{\frac{2}{3}}}}{(t - 9)^{2}}=\frac{-2t - 9}{3t^{\frac{2}{3}}(t - 9)^{2}}$.

Answer:

For $y = e^{p}(p + p\sqrt{p})$, $y'=e^{p}(p + 1+\sqrt{p}(p+\frac{3}{2}))$; for $f(t)=\frac{\sqrt[3]{t}}{t - 9}$, $f'(t)=\frac{-2t - 9}{3t^{\frac{2}{3}}(t - 9)^{2}}$