Question

hw_2.1_basic derivative rules
due sep 14, 2025 11:59 pm.

1. submit answer practice similar

attempt 1: 10 attempts remaining.
differentiate the function $g(x)=sqrt{x}+9e^{x}$
$g(x)=$

Explanation:

Step1: Recall derivative rules

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ and the rule $\frac{d}{dx}(e^x)=e^x$. Rewrite $\sqrt{x}=x^{\frac{1}{2}}$.

Step2: Differentiate each term

The derivative of $x^{\frac{1}{2}}$ is $\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$, and the derivative of $9e^x$ is $9\frac{d}{dx}(e^x)=9e^x$.

Step3: Combine the derivatives

Since $G(x)=x^{\frac{1}{2}}+9e^x$, then $G'(x)=\frac{1}{2\sqrt{x}} + 9e^x$.

Answer:

$G'(x)=\frac{1}{2\sqrt{x}}+9e^x$