1. - / 1 points differentiate the function. (h(t)=sqrt9{t}-9e^{t}) (h(t)=) 2. - / 1 points differentiate the…

Explanation:

Step1: Rewrite the functions

Rewrite $\sqrt[9]{t}=t^{\frac{1}{9}}$ and $\frac{2}{\sqrt[3]{x}} = 2x^{-\frac{1}{3}}$.

Step2: Differentiate $h(t)$

Use the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$ and $\frac{d}{dt}(e^t)=e^t$. For $h(t)=t^{\frac{1}{9}}-9e^t$, $h'(t)=\frac{1}{9}t^{\frac{1}{9}-1}-9e^t=\frac{1}{9}t^{-\frac{8}{9}}-9e^t$.

Step3: Differentiate $y$

For $y = 6e^x+2x^{-\frac{1}{3}}$, use the rule $\frac{d}{dx}(e^x)=e^x$ and $\frac{d}{dx}(x^n)=nx^{n - 1}$. So $y'=6e^x+2\times(-\frac{1}{3})x^{-\frac{1}{3}-1}=6e^x-\frac{2}{3}x^{-\frac{4}{3}}$.

Answer:

1. $h'(t)=\frac{1}{9t^{\frac{8}{9}}}-9e^t$
2. $y'=6e^x-\frac{2}{3x^{\frac{4}{3}}}$