# How do you differentiate f(x)=((18x)/(4+x^2)) using the quotient rule?

Oct 31, 2015

The derivative is $\frac{18 \left(4 - {x}^{2}\right)}{{\left(4 + {x}^{2}\right)}^{2}}$

#### Explanation:

The quotient rule states that

d/dx(f(x)/g(x)) = (f'(x)*g(x) - f(x)*g'(x))/g^2(x) color(white)(XXXXX) ( § )

Let's calculate all the quantities we need:

• Since $f \left(x\right) = 18 x$, we have that $f ' \left(x\right) = 18$.
• Since $g \left(x\right) = 4 + {x}^{2}$, we have that $g ' \left(x\right) = 2 x$.
• As for $g \left(x\right)$, we'll just write ${\left(4 + {x}^{2}\right)}^{2}$.

So, if we substitute all these functions in (§), we get

$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{18 \cdot \left(4 + {x}^{2}\right) - 18 x \cdot 2 x}{{\left(4 + {x}^{2}\right)}^{2}}$

we can easily expand the numerator into

$18 \cdot 4 + 18 {x}^{2} - 36 {x}^{2} = 72 - 18 {x}^{2} = 18 \left(4 - {x}^{2}\right)$

So, the expression becomes

$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{18 \left(4 - {x}^{2}\right)}{{\left(4 + {x}^{2}\right)}^{2}}$