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## Differential Calculus

### Course: Differential Calculus>Unit 2

Lesson 11: Quotient rule

# Quotient rule

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.2 (EK)
Introduction to the quotient rule, which tells us how to take the derivative of a quotient of functions.

## Want to join the conversation?

• I noticed that a proof is not available in this section of derivatives. Can someone explain to me the proof?
• Why is the derivative of x^2 equal to 2x?
• The derivative of a function f(x) is given by
Lim h -> 0 (f(x+h) - f(x))/h

If we have f(x) = x² then
Lim h -> 0 ((x+h)² -x²)/h =
Lim h -> 0 (x² + 2hx + h² - x²)/h
= Lim h -> 0 (2hx + h²)/h
= Lim h -> 0 2x + h
= 2x

You can also get the result from using the "power rule", discussed in this video:
You should have covered this before starting on the quotient rule.
• In this video, he gives the example of (x^2)/cos(x). But wouldn't there be an asymptote at x=pi/2? Why is it we can differentiate this function if it is not continuous?
• The function is continuous and differentiable everywhere except where cos(x)=0. The function has vertical asymptotes, and its derivative has asymptotes in the same places.
• Can someone explain to me why we sometimes can differentiate simple functions using the power rule (such as f(x) = 1/(x^2) = x^-2, thus f'(x) = -2(x)^-3) and sometimes we require the quotient rule to solve these problems? [For example, g(x) = 1/(x^2 - 1) which would leat to f'(x) = -(2x)/(x^4-1)?] Thank you!
• We can always use the power rule instead of the quotient rule. However, this isn't possible without another rule called the chain rule, so it's best to stick with the quotient rule until you learn the chain rule.

On another note, I believe you may have made a mistake in your use of the quotient rule for your g(x) function. The correct answer for g'(x) should be (x^2-2x-1)/(x^4-2x+1). I think you may have made a mistake by cancelling the (x^2-1) in the denominator with the one in the numerator. You can't do that because the one in the numerator also has a 2x being subtracted, so there aren't actually common factors to cancel.
• I think I'm a little shaking on simplifying some of these expressions. Could somebody please help me out...
d/dx √x/cos(x) =1/2√x · cos(x) - √x · -sin(x)/cos²(x)
=cos(x)+√x sin(x) / 2√x · cos²(x)
Is that right so far? The final expression = cos(x)+2xsin(x) / 2√x·cos²(x), as per Kahn Academy.
How did they get to 2xsin(x) in the numerator? I must be missing something.
• I'm beginning to think f is a lie.
• At I'm not seeing how the X^2 becomes positive
• A negative times a negative is positive.
(1 vote)
• My teacher taught us this except he said to do : (Low)(D-High) - (High)(D-Low) and then the bottom fraction squared. You have the same setup except the derivatives are switched and you have the top fraction coming first in the equation instead of the bottom one. I have a quiz on this tomorrow and now I'm just confused.