Question

differentiate the function.
y = 2e^x+\frac{5}{sqrt3{x}}
y=

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differentiate the function.
s(r)=8pi r^2
s(r)=

Explanation:

Step1: Recall derivative rules

The derivative of $e^x$ is $e^x$, and for $x^n$ the derivative is $nx^{n - 1}$. Rewrite $\frac{5}{\sqrt[3]{x}}$ as $5x^{-\frac{1}{3}}$.

Step2: Differentiate term - by - term

For the function $y = 2e^x+5x^{-\frac{1}{3}}$, the derivative of $2e^x$ is $2e^x$ (since the derivative of $e^x$ is $e^x$ and we multiply by the constant 2). The derivative of $5x^{-\frac{1}{3}}$ is $5\times(-\frac{1}{3})x^{-\frac{1}{3}- 1}=-\frac{5}{3}x^{-\frac{4}{3}}$.
So $y'=2e^x-\frac{5}{3x^{\frac{4}{3}}}$.

Step3: Differentiate $S(R) = 8\pi R^2$

Using the power rule for differentiation ($\frac{d}{dR}(aR^n)=anR^{n - 1}$, where $a = 8\pi$ and $n = 2$), we have $S'(R)=8\pi\times2R=16\pi R$.

Answer:

For $y = 2e^x+\frac{5}{\sqrt[3]{x}}$, $y'=2e^x-\frac{5}{3x^{\frac{4}{3}}}$
For $S(R)=8\pi R^2$, $S'(R)=16\pi R$