question find $k(x)$ if $k(x)=x(2x + \frac{1}{x + 3})$. provide your answer below: $k(x)=square$

Explanation:

Step1: Expand the function

First, expand (k(x)=x\left(2x + \frac{1}{x + 3}\right)=2x^{2}+\frac{x}{x + 3}).

Step2: Differentiate term - by - term

Differentiate (y = 2x^{2}) and (y=\frac{x}{x + 3}) separately. The derivative of (y = 2x^{2}) using the power rule ((x^n)^\prime=nx^{n - 1}) is (y^\prime=4x). For (y=\frac{x}{x + 3}), use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = x), (u^\prime=1), (v=x + 3), (v^\prime=1). Then ((\frac{x}{x + 3})^\prime=\frac{1\times(x + 3)-x\times1}{(x + 3)^{2}}=\frac{3}{(x + 3)^{2}}).

Step3: Combine the derivatives

(k^\prime(x)) is the sum of the derivatives of the two terms. So (k^\prime(x)=4x+\frac{3}{(x + 3)^{2}}).

Answer:

(4x+\frac{3}{(x + 3)^{2}})