# How do you differentiate f(x)=x/(x-4)^2 using the quotient rule?

Dec 21, 2016

See below.

#### Explanation:

The quotient rule is given by

In this case, $f \left(x\right) = x$ and $g \left(x\right) = {\left(x - 4\right)}^{2}$. We will also make use of the chain rule.

As indicated, we will first take the derivative of $f \left(x\right)$ and multiply this by $g \left(x\right)$, which we leave as is.

$f ' \left(x\right) = 1$

$g \left(x\right) f ' \left(x\right) = {\left(x - 4\right)}^{2}$

From this, we subtract the product of $g ' \left(x\right)$ and $f \left(x\right)$, which we leave as is. We will use the chain rule to differentiate $g \left(x\right)$.

$g ' \left(x\right) = 2 \left(x - 4\right) \left(1\right)$

$f \left(x\right) g ' \left(x\right) = 2 x \left(x - 4\right)$

And, ${\left[g \left(x\right)\right]}^{2}$ is ${\left[{\left(x - 4\right)}^{2}\right]}^{2} = {\left(x - 4\right)}^{4}$

The derivative is then given as:

$f ' \left(x\right) = \frac{{\left(x - 4\right)}^{2} - 2 x \left(x - 4\right)}{x - 4} ^ 4$

Which simplifies to

$\frac{\left(x - 4\right) - 2 x}{x - 4} ^ 3$

$\implies \frac{- x - 4}{x - 4} ^ 3$

$\implies f ' \left(x\right) = - \frac{x + 4}{x - 4} ^ 3$