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# Derivative rules review

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)
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

## 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

## 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

## 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

## 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 join the conversation?

• Can somebody explain the chain rule in an "easy-to-understand" way? Thanks :)
• 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)
• 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) ?
• 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.
• 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.
• 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.
• The variable power rule is missing: a^x = ln(a) a^x and differentiation rules for logarithms
• 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).