Question

find the derivative of the following function.
s(w)=\frac{9w + 2}{sqrt{w}+6}
the derivative of s(w)=\frac{9w + 2}{sqrt{w}+6} is square.

Explanation:

Step1: Identify quotient - rule

The quotient - rule states that if $s(w)=\frac{u(w)}{v(w)}$, then $s^\prime(w)=\frac{u^\prime(w)v(w)-u(w)v^\prime(w)}{v(w)^2}$. Here, $u(w)=9w + 2$ and $v(w)=\sqrt{w}+6=w^{\frac{1}{2}}+6$.

Step2: Find $u^\prime(w)$

Differentiate $u(w)=9w + 2$ with respect to $w$. Using the power - rule $\frac{d}{dw}(aw + b)=a$ (where $a = 9$ and $b = 2$), we get $u^\prime(w)=9$.

Step3: Find $v^\prime(w)$

Differentiate $v(w)=w^{\frac{1}{2}}+6$ with respect to $w$. Using the power - rule $\frac{d}{dw}(w^n)=nw^{n - 1}$, we have $v^\prime(w)=\frac{1}{2}w^{-\frac{1}{2}}=\frac{1}{2\sqrt{w}}$.

Step4: Apply the quotient - rule

\[

$$\begin{align*} s^\prime(w)&=\frac{u^\prime(w)v(w)-u(w)v^\prime(w)}{v(w)^2}\\ &=\frac{9(\sqrt{w}+6)-(9w + 2)\frac{1}{2\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{9\sqrt{w}+54-\frac{9w}{2\sqrt{w}}-\frac{2}{2\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{9\sqrt{w}+54-\frac{9\sqrt{w}}{2}-\frac{1}{\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{\frac{18\sqrt{w}-9\sqrt{w}}{2}+54-\frac{1}{\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{\frac{9\sqrt{w}}{2}+54-\frac{1}{\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{\frac{9w + 108\sqrt{w}-2}{2\sqrt{w}}}{(\sqrt{w}+6)^2}\\ &=\frac{9w + 108\sqrt{w}-2}{2\sqrt{w}(\sqrt{w}+6)^2} \end{align*}$$

\]

Answer:

$\frac{9w + 108\sqrt{w}-2}{2\sqrt{w}(\sqrt{w}+6)^2}$