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 $y^\prime=u^\prime v + uv^\prime$. Let $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=\ln(x - 3)$.

Step2: Differentiate $u$

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

Step3: Differentiate $v$

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

Step4: Apply product - rule formula

$g^\prime(x)=u^\prime v+uv^\prime=\frac{1}{2\sqrt{x}}\ln(x - 3)+\sqrt{x}\cdot\frac{1}{x - 3}$.

Answer:

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