- Derivative of aˣ (for any positive base a)
- Derivative of logₐx (for any positive base a≠1)
- Derivatives of aˣ and logₐx
- Worked example: Derivative of 7^(x²-x) using the chain rule
- Worked example: Derivative of log₄(x²+x) using the chain rule
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Worked example: Derivative of ∜(x³+4x²+7) using the chain rule
- Chain rule capstone
- Proving the chain rule
- Derivative rules review
Derivative rules review
Review all the common derivative rules (including Power, Product, and Chain rules).
Basic differentiation rules
Constant multiple rule:
Want to learn more about the basic differentiation rules? Check out this video.
Want to learn more about the Power rule? Check out this video.
Want to learn more about the Product rule? Check out this video.
Want to learn more about the Quotient rule? Check out this video.
Want to learn more about the Chain rule? Check out this video.
Want to join the conversation?
- Can somebody explain the chain rule in an "easy-to-understand" way? Thanks :)(10 votes)
- 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:
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)(26 votes)
- 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)
- 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)
- 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)
- 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.(15 votes)
- The variable power rule is missing:
a^x = ln(a) a^xand differentiation rules for logarithms(4 votes)
- 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)
- For the equation
y = x(x - 4) ^ 3you 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
g'(x)=3(x-4)^2, so if you plug it in you get
1 * (x-4)^3 + 3(x-4)^2 * xwhich I simplified to
(x - 4)^2 (x-4+3x)or
Hope this help!(1 vote)
- This has gotta be all for Calculus AB, isn't it?(1 vote)
- Yes. Everything until the application of integrals is AB Calc. Parametric equations and infinite sums are BC Calc(2 votes)