Question

use the quotient rule to find the derivative of the function.
f(x)=\frac{x^{2}}{4sqrt{x}+1}
f(x)=square

Explanation:

Step1: Recall the Quotient Rule

The Quotient Rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = x^{2}$, so $u^{\prime}(x)=2x$; $v(x)=4\sqrt{x}+1 = 4x^{\frac{1}{2}}+1$, and $v^{\prime}(x)=4\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{2}{\sqrt{x}}$.

Step2: Substitute into the Quotient Rule

\[

$$\begin{align*} f^{\prime}(x)&=\frac{2x(4\sqrt{x}+1)-x^{2}\times\frac{2}{\sqrt{x}}}{(4\sqrt{x}+1)^{2}}\\ &=\frac{2x(4x^{\frac{1}{2}} + 1)-2x^{2}\times x^{-\frac{1}{2}}}{(4x^{\frac{1}{2}}+1)^{2}}\\ &=\frac{8x^{\frac{3}{2}}+2x - 2x^{\frac{3}{2}}}{(4x^{\frac{1}{2}}+1)^{2}}\\ &=\frac{6x^{\frac{3}{2}}+2x}{(4x^{\frac{1}{2}}+1)^{2}} \end{align*}$$

\]

Answer:

$\frac{6x^{\frac{3}{2}}+2x}{(4\sqrt{x}+1)^{2}}$