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Derivative rules review

Review all the common derivative rules (including Power, Product, and Chain rules).

Basic differentiation rules

Constant rule: start fraction, d, divided by, d, x, end fraction, left parenthesis, k, right parenthesis, equals, 0
Sum rule: start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, plus, g, left parenthesis, x, right parenthesis, close bracket, equals, f, prime, left parenthesis, x, right parenthesis, plus, g, prime, left parenthesis, x, right parenthesis
Difference rule: start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, minus, g, left parenthesis, x, right parenthesis, close bracket, equals, f, prime, left parenthesis, x, right parenthesis, minus, g, prime, left parenthesis, x, right parenthesis
Constant multiple rule: start fraction, d, divided by, d, x, end fraction, open bracket, k, f, left parenthesis, x, right parenthesis, close bracket, equals, k, f, prime, left parenthesis, x, right parenthesis
Want to learn more about the basic differentiation rules? Check out this video.

Power rule

start fraction, d, divided by, d, x, end fraction, left parenthesis, x, start superscript, n, end superscript, right parenthesis, equals, n, dot, x, start superscript, n, minus, 1, end superscript
Want to learn more about the Power rule? Check out this video.

Product rule

start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, close bracket, equals, f, prime, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, plus, f, left parenthesis, x, right parenthesis, g, prime, left parenthesis, x, right parenthesis
Want to learn more about the Product rule? Check out this video.

Quotient rule

start fraction, d, divided by, d, x, end fraction, left parenthesis, start fraction, f, left parenthesis, x, right parenthesis, divided by, g, left parenthesis, x, right parenthesis, end fraction, right parenthesis, equals, start fraction, f, prime, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, minus, f, left parenthesis, x, right parenthesis, g, prime, left parenthesis, x, right parenthesis, divided by, open bracket, g, left parenthesis, x, right parenthesis, close bracket, squared, end fraction
Want to learn more about the Quotient rule? Check out this video.

Chain rule

start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, close bracket, equals, f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis
Want to learn more about the Chain rule? Check out this video.

Want to join the conversation?

  • hopper cool style avatar for user Moksha
    Can somebody explain the chain rule in an "easy-to-understand" way? Thanks :)
    (10 votes)
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    • aqualine seedling style avatar for user Seongjoo
      Consider a function (x+1)^2.
      We can see that the function has two parts, the enclosing part
      (outside: ( )^2) and the enclosed part (inside: x+1).

      To do the chain rule you first take the derivative of the outside as if you would normally (disregarding the inner parts), then you add the inside back into the derivative of the outside.
      Afterwards, you take the derivative of the inside part and multiply that with the part you found previously.

      So to continue the example:
      d/dx[(x+1)^2]
      1. Find the derivative of the outside:
      Consider the outside ( )^2 as x^2 and find the derivative
      as d/dx x^2 = 2x
      the outside portion = 2( )
      2. Add the inside into the parenthesis:
      2( ) = 2(x+1)
      3. Find the derivative of the inside and multiply:
      as d/dx [x+1] = 1
      1*2(x+1) = 2(x+1).

      Thus, d/dx[(x+1)^2] = 2(x+1)
      (31 votes)
  • spunky sam blue style avatar for user abu
    In terms of the AP exam: Are proofs even needed? I know that you need to get a good conceptual idea, but my math teacher claims no one ever uses proofs. If they want, they can take a course on proofs in college, but teachers would lose kids interest with all these proofs. I'm 99 percent sure proofs are not required on the AP exam? Someone correct me if I'm wrong.
    (0 votes)
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    • aqualine ultimate style avatar for user Mark Geary
      Proofs are NOT specifically needed for the AP exam. However, working through proofs can be illuminating into how functions (and their derivations) work. Practically speaking, the more you are able to manipulate functions algebraically and trigonometrically, the better you will be able to work with function problems at the AP (or any) exam. It's a matter of familiarity, fluency, and functionality. You don't have to be able to do a proof at the exam, but if you've done a lot of proofs beforehand, you'll likely score better at the exam.
      (20 votes)
  • blobby green style avatar for user foo bar
    The variable power rule is missing: a^x = ln(a) a^x and differentiation rules for logarithms
    (7 votes)
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  • aqualine ultimate style avatar for user Anurag
    In quotient rule, can't we write the derivative of f(x)/g(x) as
    f(x).g'(x)^-1 + g(x)^-1.f'(x) ?
    (2 votes)
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    • leaf yellow style avatar for user Howard Bradley
      No, because d/dx (1/g(x)) is not 1/g'(x). You have to apply the chain rule. It should be -1/(g(x))² · g'(x)
      So we'd have (using the product rule and the chain rule):
      d/dx f(x) · 1/g(x) = 1/g(x) · f'(x) + f(x) · -1/(g(x))² · g'(x)

      Which, with a bit of manipulation, can be made to look like the familiar quotient rule for differentiation.
      (9 votes)
  • hopper cool style avatar for user darknessbright
    we could also use the power rule to prove that d/dx[x]=1
    (3 votes)
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  • piceratops sapling style avatar for user DRippy_CHexxx
    Can you apply chainrule for questions like: ddx of [f(g(h(x)))]?
    (2 votes)
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  • male robot hal style avatar for user James
    This has gotta be all for Calculus AB, isn't it?
    (2 votes)
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  • blobby green style avatar for user Melia Jane
    Hey Great Learners. x(x-4)^3 becomes (x-4)^2(4x-4) How is this when using the chain rule? I get 3x(x-4)^2(1).
    Please and Thanks!
    (2 votes)
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    • leaf grey style avatar for user Chace Caven
      For the equation y = x(x - 4) ^ 3 you have to use the product rule AND the chain rule since it's x * (x-4)^3. If you're using the formula d/dx f(x)g(x) = f'(x)g(x) + g'(x)f(x) then your f(x)=x and g(x)=(x-4)^3 so f'(x)=1 and g'(x)=3(x-4)^2, so if you plug it in you get 1 * (x-4)^3 + 3(x-4)^2 * x which I simplified to (x - 4)^2 (x-4+3x) or (x-4)^2(4x-4).

      Hope this help!
      (1 vote)