Question

1. differentiate $g(x)=sqrt{x}ln(x - 3)$ answer: $g(x)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=\ln(x - 3)$.

Step2: Differentiate $u$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $u = x^{\frac{1}{2}}$, we have $u^\prime=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}$.

Step3: Differentiate $v$

Using the chain - rule, if $v=\ln(u)$ and $u=x - 3$, then $\frac{dv}{dx}=\frac{1}{u}\cdot\frac{du}{dx}$. Since $\frac{du}{dx}=1$ and $u=x - 3$, $v^\prime=\frac{1}{x - 3}$.

Step4: Calculate $g^\prime(x)$

By the product - rule $g^\prime(x)=u^\prime v+uv^\prime$. Substitute $u = \sqrt{x}$, $u^\prime=\frac{1}{2\sqrt{x}}$, $v=\ln(x - 3)$ and $v^\prime=\frac{1}{x - 3}$ into the formula:
\[

$$\begin{align*} g^\prime(x)&=\frac{1}{2\sqrt{x}}\ln(x - 3)+\sqrt{x}\cdot\frac{1}{x - 3}\\ &=\frac{\ln(x - 3)}{2\sqrt{x}}+\frac{\sqrt{x}}{x - 3} \end{align*}$$

\]

Answer:

$\frac{\ln(x - 3)}{2\sqrt{x}}+\frac{\sqrt{x}}{x - 3}$