Question

hw 14 - derivatives of logs section 2.9: problem 3 (1 point) find $\frac{dy}{dx}$ when $y = ln(15 + 5sqrt{x})$. $\frac{dy}{dx}=square$

Explanation:

Step1: Identify the outer - inner functions

Let $u = 15 + 5\sqrt{x}=15 + 5x^{\frac{1}{2}}$, and $y=\ln(u)$.

Step2: Find the derivative of the outer function

The derivative of $y = \ln(u)$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{u}$.

Step3: Find the derivative of the inner function

The derivative of $u = 15+5x^{\frac{1}{2}}$ with respect to $x$ is $\frac{du}{dx}=0 + 5\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{5}{2\sqrt{x}}$.

Step4: Use the chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=\frac{5}{2\sqrt{x}}$ into the chain - rule formula. Since $u = 15 + 5\sqrt{x}$, we have $\frac{dy}{dx}=\frac{1}{15 + 5\sqrt{x}}\cdot\frac{5}{2\sqrt{x}}$.

Step5: Simplify the expression

$\frac{dy}{dx}=\frac{5}{2\sqrt{x}(15 + 5\sqrt{x})}=\frac{5}{30\sqrt{x}+10x}=\frac{1}{6\sqrt{x}+2x}$.

Answer:

$\frac{1}{6\sqrt{x}+2x}$