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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 3

Lesson 8: Rational functions differentiation

# Differentiating rational functions review

Review your rational function differentiation skills and use them to solve problems.

## How do I differentiate rational functions?

To differentiate rational functions, all you need is to know how to differentiate polynomials and the quotient rule. Let's differentiate $\frac{{x}^{2}-2}{x-1}$ for example.
$\begin{array}{rl}& =\frac{d}{dx}\left(\frac{{x}^{2}-2}{x-1}\right)\\ \\ & =\frac{\left(x-1\right)\frac{d}{dx}\left({x}^{2}-2\right)-\left({x}^{2}-2\right)\frac{d}{dx}\left(x-1\right)}{\left(x-1{\right)}^{2}}& & \text{Quotient rule}\\ \\ & =\frac{\left(x-1\right)\left(2x\right)-\left({x}^{2}-2\right)\left(1\right)}{\left(x-1{\right)}^{2}}\\ \\ & =\frac{2{x}^{2}-2x-{x}^{2}+2}{\left(x-1{\right)}^{2}}\\ \\ & =\frac{{x}^{2}-2x+2}{\left(x-1{\right)}^{2}}\end{array}$

## Practice set 1: Differentiate rational functions

Problem 1.1
$g\left(x\right)=\frac{x-5}{{x}^{2}+1}$
${g}^{\prime }\left(x\right)=$

Want to try more problems like this? Check out this exercise.

## Practice set 2: Evaluate rational functions derivatives

Problem 2.1
$f\left(x\right)=\frac{3{x}^{2}+1}{x+2}$
${f}^{\prime }\left(-3\right)=?$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• For f(x) = x • |x|, at what point is f(x) differentials? • For x * |x| it should be differentiable everywhere (you can simply check the graph too, though this is not rigorous/formal). We have the definition:

x * x for x >= 0
x * -x for x < 0

This gives us the piecewise functions x^2 and - x^2 (notice the - is NOT in parenthesis, so yes, it is -1 * (x^2). The graph tells us this is true, It's like a normal x^2 parabola to the right, but a negative parabola to the left (opens downward).

To check for continuity we ensure the lim of the left and right sides are equal. We can see when we approach values from negative infinity we have L = 0, and when we approach from positive infinity we get Lim = 0, so the limits are equal, and the function is continuous.

For differentiability we can differentiate each function, and check and see if their limits would be the same:

We have -2x and 2x as x approaches 0, and indeed their limits are L = 0 for each, therefore we can say that f(x) = x * |x| is continuous AND differentiable for ALL values of x.

And yes |x| is defined for 0, I think what kubleeka is referring to is the normal |x| function. This function is continuous, but NOT DIFFERENTIABLE, because we get (when differentiating) -1 = 1 for the limits, basically when we come from the left side from negative infinity toward 0 we have a negative slope (-1), BUT when we come in from the right side from positive infinity we have a positive slope (1), so we have a cusp, or sharp curve. As you know we can't differentiate in a few cases, like jump disconuities, cusps/sharp turns, etc.

When in doubt:

1. Check the limit for the function (or functions) and ensure their limits are equal, this ensure continuity.
2. Differentiate the function/functions and ensure THEIR limits are equal, this ensure there are no disconuities for differentiability and it will be differentiable (no cusps, no jumps, etc).
• what is the answer to$-6 \cdot f(-3) - 5 \cdot g(-7) =$ 