# How do you find the derivative of f(x) = (x^2-4)/(x-1)?

Jun 6, 2016

In a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, the derivative is given by $\frac{\left(g ' \left(x\right) \times h \left(x\right)\right) - \left(g \left(x\right) \times h ' \left(x\right)\right)}{h \left(x\right)} ^ 2$

#### Explanation:

We must determine the derivatives of the numerator and of the denominator,

Numerator:

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

By the power rule, $g ' \left(x\right) = 2 x$.

Denominator:

Let $h \left(x\right) = x - 1$

By the power rule, $h ' \left(x\right) = 1$
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We can now differentiate with the quotient rule.

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

$f ' \left(x\right) = \frac{\left(2 x \left(x - 1\right)\right) - \left(\left({x}^{2} - 4\right) 1\right)}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{2} - 2 x - {x}^{2} + 4}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} - 2 x + 4}{{x}^{2} - 2 x + 1}$

Thus, the derivative of $f \left(x\right) = \frac{{x}^{2} - 4}{x - 1}$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 2 x + 4}{{x}^{2} - 2 x + 1}$

Hopefully this helps!