Question

differentiate the function with respect to the independent variable.
f(x)=\ln\frac{x^{3}-125}{x^{2}-25}
f(x)=\square

Explanation:

Step1: Use logarithm properties

Use $\ln\frac{a}{b}=\ln a-\ln b$. So $f(x)=\ln(x^{3} - 125)-\ln(x^{2}-25)$.

Step2: Apply the chain rule

The derivative of $\ln u$ is $\frac{u'}{u}$.
For $y = \ln(x^{3}-125)$, let $u=x^{3}-125$, then $y'=\frac{3x^{2}}{x^{3}-125}$.
For $y=\ln(x^{2}-25)$, let $u = x^{2}-25$, then $y'=\frac{2x}{x^{2}-25}$.

Step3: Calculate $f'(x)$

$f'(x)=\frac{3x^{2}}{x^{3}-125}-\frac{2x}{x^{2}-25}$.
Find a common denominator $(x^{3}-125)(x^{2}-25)$:

$$
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$$

Answer:

$\frac{3x^{2}}{x^{3}-125}-\frac{2x}{x^{2}-25}$ (or $\frac{x(x^{3}-75x + 250)}{(x^{3}-125)(x^{2}-25)}$ after simplification)