find the derivative of y with respect to x.\ny = (2x + 3)^x\n\frac{dy}{dx}=square\n(use parentheses to…

Explanation:

Step1: Take natural - log of both sides

$\ln y=x\ln(2x + 3)$

Step2: Differentiate both sides with respect to x

Using the product rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = x$ and $v=\ln(2x + 3)$. The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}\frac{dy}{dx}$, the derivative of $x$ is $1$, and the derivative of $\ln(2x + 3)$ using the chain - rule is $\frac{2}{2x+3}$. $\frac{1}{y}\frac{dy}{dx}=\ln(2x + 3)+x\cdot\frac{2}{2x + 3}$

Step3: Solve for $\frac{dy}{dx}$

Multiply both sides by $y=(2x + 3)^x$. $\frac{dy}{dx}=(2x + 3)^x\left(\ln(2x + 3)+\frac{2x}{2x + 3}\right)$

Answer:

$(2x + 3)^x\left(\ln(2x + 3)+\frac{2x}{2x + 3}\right)$