If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎ Calculus AB (2017 edition)

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

Lesson 8: Rational functions differentiation

# Differentiating rational functions

Let's dive into the differentiation of the rational function (5-3x)/(x²+3x) using the Quotient Rule. By identifying the numerator and denominator as separate functions, we apply the Quotient Rule to find the derivative, simplifying the expression for a clear understanding of the process. This approach can be applied to differentiate any other rational function as well.

## Want to join the conversation?

• Why does sal never simplify the ((x^2)+3x)^2 in the denominator?
• Because simplifying the denominator can lead to unnecessary algebraic mistakes and a more cluttered fraction. My calculus teacher doesn't even have us simplify our answers to problems because even the smallest mistake in solving for the answer can lead to an incorrect answer. Leaving the denominator in this form ensures the mathematician that they don't have to worry about small mistakes
• What does it mean for the derivative to exist at every point on the domain of it's function? It seems like an abstract thing to visualize.
Can someone explain?
• It means that for all real numbers (in the domain) the function has a derivative.

For this to be true the function must be defined, continuous and differentiable at all points.

In other words, there are no discontinuities, no corners AND no vertical tangents.

ADDENDUM: An example of the importance of the last condition is the function f(x) = x^(1/3) — this function is defined and continuous for all real numbers, but at x=0 it has a vertical tangent and thus is not differentiable.
• Why would you need to do partial fraction if you have quotient rule?
• You wouldn't, for differentiation. However, for solving some integrals, partial fraction decomposition is the only way to make progress.
• In this lesson Sal uses dy/dx where he only used and explained d/dx in the past. What is the difference between the two, if any?
• d/dx - Derivative of [blank] with respect to x. In and of itself, it's meaningless, like a plus sign operator by itself.
dy/dx - Derivative of y with respect to x. An actual quantity or function (if y is differentiable).
• take d/dx[(2x + 3)/(3x^2 - 4)] at x=-1. Say f(x) = 2x + 3 and g(x) = 3x^2 - 4. So that's [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2.
Plug in the expressions to get d/dx(2x + 3)(3x^2 -4) - (2x+3)d/dx(3x^2 - 4), all divided by (3x^2 - 4)^2.
If you do the math, the derivative is 4.
But what happens if, when taking the derivative of a constant, you made it 1 instead of 0? So d/dx(2x + 3) = 2 + 1 = 3.
Do that through the whole problem. The final answer is still 4. And this is true of not just this expression. How is this happening?
• The derivative of a constant is 0 not 1.

If you are getting the same answer using both methods that is just a coincidence — it is not universally true.

I suggest trying this with more examples — you will quickly find that it usually does not work!
• I rewrote f(x) as (5 - 3x)*(x^2 + 3x)^(-1) and used the product rule. The answer was different. Why can't I rewrite the function to be a product and then find its derivative? Dividing by something is the same as multiplying by its reciprocal
• You must have made an error in your math somewhere. You should get the same answer when you use the product rule.

f(x)=(5-3x)*(x^2+3x)^(-1)
f'(x)=(5-3x)'*(x^2+3x)^(-1)+(5-3x)*((x^2+3x)^(-1))'
f'(x)=-3*(x^2+3x)^(-1)-(5-3x)*(2x+3)*(x^2+3x)^(-2)
f'(x)=(-3*(x^2+3x)-(5-3x)(2x+3))*(x^2+3x)^(-2)
f'(x)=(-3*(x^2+3x)-(5-3x)(2x+3))/(x^2+3x)^2
f'(x)=(-3x^2-9x-10x-15+6x^2+9x)/(x^2+3x)^2
f'(x)=(3x^2-10x-15)/(x^2+3x)^2