differentiate. g(x) = 5e^x√x g(x) = need help? read it submit answer

Explanation:

Step1: Recall product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, let $u = 5e^{x}$ and $v=\sqrt{x}=x^{\frac{1}{2}}$.

Step2: Differentiate $u$

The derivative of $u = 5e^{x}$ with respect to $x$ is $u'=\frac{d}{dx}(5e^{x}) = 5e^{x}$.

Step3: Differentiate $v$

The derivative of $v=x^{\frac{1}{2}}$ with respect to $x$ is $v'=\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$.

Step4: Apply product - rule

$g'(x)=u'v + uv'$. Substitute $u = 5e^{x}$, $u' = 5e^{x}$, $v=x^{\frac{1}{2}}$, and $v'=\frac{1}{2}x^{-\frac{1}{2}}$ into the formula: [ \begin{align*} g'(x)&=5e^{x}\cdot x^{\frac{1}{2}}+5e^{x}\cdot\frac{1}{2}x^{-\frac{1}{2}}\ &=5e^{x}\sqrt{x}+\frac{5e^{x}}{2\sqrt{x}}\ &=\frac{10xe^{x}+ 5e^{x}}{2\sqrt{x}}\ &=\frac{5e^{x}(2x + 1)}{2\sqrt{x}} \end{align*} ]

Answer:

$\frac{5e^{x}(2x + 1)}{2\sqrt{x}}$