Question

question
what is the first derivative of y(x) = log₇(2x³ + 3x² + 5)?
select the correct answer below:
○ $\frac{6x^{2}+6x}{(ln 7)(2x^{3}+3x^{2}+5)}$
○ $\frac{6x^{2}+6x}{2x^{3}+3x^{2}+5}$
○ $\frac{1}{(ln 7)(2x^{3}+3x^{2}+5)}$
○ $\frac{(ln 7)(6x^{2}+6x)}{2x^{3}+3x^{2}+5}$
○ log₇(6x² + 6x)

Explanation:

Step1: Use change - of - base formula

$y(x)=\frac{\ln(2x^{3}+3x^{2}+5)}{\ln7}$

Step2: Apply chain - rule

$y^\prime(x)=\frac{1}{\ln7}\cdot\frac{6x^{2}+6x}{2x^{3}+3x^{2}+5}=\frac{6x^{2}+6x}{(\ln7)(2x^{3}+3x^{2}+5)}$

Answer:

$\frac{6x^{2}+6x}{(\ln7)(2x^{3}+3x^{2}+5)}$