# What is the first derivative and critical numbers of y= (x^4)/(x^4-1)?

Nov 21, 2017

Critical points are at $x = - 1 , x = 0 , x = 1$

#### Explanation:

This derivative can of course be computed by using the quotient rule. However, that tends to be somewhat cumbersome, so we can be clever. If we use a bit of algebraic long division, we can get away with a simple chain rule application.

${x}^{4} / \left({x}^{4} - 1\right) = 1 + \frac{1}{{x}^{4} - 1} = 1 + {\left({x}^{4} - 1\right)}^{-} 1$ (using algebraic long division)

$\frac{d}{\mathrm{dx}} \left(1 + {\left({x}^{4} - 1\right)}^{-} 1\right) =$

Let $u = {x}^{4} - 1$:

$= \frac{d}{\mathrm{du}} \left(1 + {u}^{-} 1\right) \frac{d}{\mathrm{dx}} \left({x}^{4} - 1\right) =$ (using the chain rule)

$= - {u}^{-} 2 \cdot 4 {x}^{3} =$ (by the power rule)

Resubstitute:

$= - {\left({x}^{4} - 1\right)}^{-} 2 \cdot 4 {x}^{3} = - \frac{4 {x}^{3}}{{\left({x}^{4} - 1\right)}^{2}}$

To solve for the critical points, we simply set the derivative expression equal to 0 and solve for x, and also find any x's where the function is undefined:

$- \frac{4 {x}^{3}}{{\left({x}^{4} - 1\right)}^{2}} = 0$

When we have an expression of something divided by something else equals 0, we can deduce that the top expression must be 0, since if the bottom were to be 0, the value would be undefined.

$- 4 {x}^{3} = 0$

${x}^{3} = 0$

$x = \sqrt[3]{0}$

$x = 0$

So, we know that $x = 0$ is the only point that the derivative of the function equals zero. However, we also have to consider what would make this function undefined. We said before that if the denominator would be equal to 0, then the expression would be undefined, so let's set the denominator equal to 0:

${\left({x}^{4} - 1\right)}^{2} = 0$

${x}^{4} - 1 = \sqrt{0} = 0$

${x}^{4} = 1$

$x = \pm \sqrt[4]{1}$

$x = \pm 1$

So, if $x = 1$ or $x = - 1$, the denominator would equal to zero, which would make the function undefined, so we know that these two points also are critical points on the graph.