Find the derivative of the given function.
$h\left(x\right)=4^{{(\sqrt{x}+3x)}^{\frac{5}{4}}}$

• A. $\frac54\left(\sqrt{x}+3x\right)\left(\frac{1}{2\sqrt{x}}+3\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• B. $\frac{5\ln\left(4\right)}{4}\left(\sqrt{x}+3x\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• C. $\frac54\cdot4\left(\sqrt{x}+3x\right)^{\frac54}\cdot\ln4\cdot\left(\sqrt{x}+3x\right)^{\frac14}\cdot\left(\frac{1}{2\sqrt{x}}+3\right)$
• D. $\left(\sqrt{x}+3x\right)^{\frac54}\cdot4\left(\sqrt{x}+3x\right)^{\frac14}$

Answer: (C)

Verified step by step guidance

1. Identify the function to differentiate: \( h(x) = 4(\sqrt{x} + 3x)^{\frac{5}{4}} \). This is a composite function, so we will use the chain rule.
2. Apply the chain rule: If \( y = (u(x))^n \), then \( \frac{dy}{dx} = n(u(x))^{n-1} \cdot \frac{du}{dx} \). Here, \( u(x) = \sqrt{x} + 3x \) and \( n = \frac{5}{4} \).
3. Differentiate \( u(x) = \sqrt{x} + 3x \): \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} + 3 \).
4. Substitute back into the chain rule: \( \frac{dh}{dx} = 4 \cdot \frac{5}{4} (\sqrt{x} + 3x)^{\frac{1}{4}} \cdot (\frac{1}{2\sqrt{x}} + 3) \).
5. Simplify the expression: \( \frac{dh}{dx} = 5 (\sqrt{x} + 3x)^{\frac{1}{4}} \cdot (\frac{1}{2\sqrt{x}} + 3) \).