find the derivative of the function f(x)=ln(2x + 5)(3x^2+2).\nf(x)=□

Explanation:

Step1: Use the property of logarithms

First, use $\ln(ab)=\ln a+\ln b$. So, $f(x)=\ln(2x + 5)+\ln(3x^{2}+2)$.

Step2: Differentiate $\ln(2x + 5)$

The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$. For $u = 2x+5$, $u'=2$. So the derivative of $\ln(2x + 5)$ is $\frac{2}{2x + 5}$.

Step3: Differentiate $\ln(3x^{2}+2)$

For $u = 3x^{2}+2$, $u' = 6x$. So the derivative of $\ln(3x^{2}+2)$ is $\frac{6x}{3x^{2}+2}$.

Step4: Find $f'(x)$

By the sum - rule of differentiation, $f'(x)$ is the sum of the derivatives of the two terms. So $f'(x)=\frac{2}{2x + 5}+\frac{6x}{3x^{2}+2}$.

Answer:

$\frac{2}{2x + 5}+\frac{6x}{3x^{2}+2}$