Question

question
find the derivative of $h(x)=left(\frac{x}{2 - 2x}
ight)^2$.
provide your answer below:
h(x)=□

Explanation:

Step1: Rewrite the function

Let $u(x)=\frac{x}{2 - 2x}$, so $h(x)=u^{2}(x)$. First, rewrite $u(x)=\frac{x}{2(1 - x)}=\frac{1}{2}\cdot\frac{x}{1 - x}$.

Step2: Use the quotient - rule to find $u'(x)$

The quotient - rule states that if $y=\frac{f(x)}{g(x)}$, then $y'=\frac{f'(x)g(x)-f(x)g'(x)}{g^{2}(x)}$. Here, $f(x)=x$, $f'(x)=1$, $g(x)=1 - x$, $g'(x)=-1$. So $u'(x)=\frac{1}{2}\cdot\frac{1\cdot(1 - x)-x\cdot(-1)}{(1 - x)^{2}}=\frac{1}{2}\cdot\frac{1 - x + x}{(1 - x)^{2}}=\frac{1}{2(1 - x)^{2}}$.

Step3: Use the chain - rule to find $h'(x)$

The chain - rule states that if $y = f(u)$ and $u = u(x)$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $h(x)=u^{2}(x)$, $\frac{dh}{du} = 2u$. Then $h'(x)=2u(x)\cdot u'(x)$. Substitute $u(x)=\frac{x}{2(1 - x)}$ and $u'(x)=\frac{1}{2(1 - x)^{2}}$ into the formula:
\[

$$\begin{align*} h'(x)&=2\cdot\frac{x}{2(1 - x)}\cdot\frac{1}{2(1 - x)^{2}}\\ &=\frac{x}{2(1 - x)^{3}} \end{align*}$$

\]

Answer:

$\frac{x}{2(1 - x)^{3}}$