Question

find the derivative of the following function
g(x)=\frac{4x + 1}{sqrt{x}-8}
the derivative of (g(x)=\frac{4x + 1}{sqrt{x}-8}) is (square)

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}$. Here, $u(x)=4x + 1$, so $u^\prime(x)=4$, and $v(x)=\sqrt{x}-8=x^{\frac{1}{2}}-8$, so $v^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}$.

Step2: Substitute into quotient - rule formula

\[

$$\begin{align*} g^\prime(x)&=\frac{4(\sqrt{x}-8)-(4x + 1)\frac{1}{2\sqrt{x}}}{(\sqrt{x}-8)^2}\\ &=\frac{4\sqrt{x}-32-\frac{4x}{2\sqrt{x}}-\frac{1}{2\sqrt{x}}}{(\sqrt{x}-8)^2}\\ &=\frac{4\sqrt{x}-32 - 2\sqrt{x}-\frac{1}{2\sqrt{x}}}{(\sqrt{x}-8)^2}\\ &=\frac{2\sqrt{x}-32-\frac{1}{2\sqrt{x}}}{(\sqrt{x}-8)^2}\\ &=\frac{\frac{4x-64\sqrt{x}-1}{2\sqrt{x}}}{(\sqrt{x}-8)^2}\\ &=\frac{4x-64\sqrt{x}-1}{2\sqrt{x}(\sqrt{x}-8)^2} \end{align*}$$

\]

Answer:

$\frac{4x - 64\sqrt{x}-1}{2\sqrt{x}(\sqrt{x}-8)^2}$