Question

differentiate (f(x)=ln(sqrt{(4x - 6)(5x + 5)})).
(f(x)=)

Explanation:

Step1: Simplify the function

First, use the property $\ln(\sqrt{a})=\frac{1}{2}\ln(a)$ and $\ln(ab)=\ln(a)+\ln(b)$. So $f(x)=\frac{1}{2}\ln((4x - 6)(5x + 5))=\frac{1}{2}(\ln(4x - 6)+\ln(5x + 5))$.

Step2: Differentiate term - by - term

The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$ by the chain - rule. For $u = 4x-6$, $u'=4$; for $u = 5x + 5$, $u'=5$.
The derivative of $\frac{1}{2}\ln(4x - 6)$ is $\frac{1}{2}\times\frac{4}{4x - 6}=\frac{2}{4x - 6}$, and the derivative of $\frac{1}{2}\ln(5x + 5)$ is $\frac{1}{2}\times\frac{5}{5x + 5}=\frac{5}{2(5x + 5)}$.

Step3: Combine the derivatives

$f'(x)=\frac{2}{4x - 6}+\frac{5}{2(5x + 5)}$. Simplify the fractions: $\frac{2}{4x - 6}=\frac{1}{2x - 3}$ and $\frac{5}{2(5x + 5)}=\frac{1}{2(x + 1)}$. Then find a common denominator $(2x - 3)\times2(x + 1)=4x^{2}-2x - 6$.
$f'(x)=\frac{2(x + 1)+(2x - 3)}{4x^{2}-2x - 6}=\frac{2x+2 + 2x-3}{4x^{2}-2x - 6}=\frac{4x - 1}{4x^{2}-2x - 6}$.

Answer:

$\frac{4x - 1}{4x^{2}-2x - 6}$