Review all the common derivative rules (including Power, Product, and Chain rules).
Log in Moksha 4 years agoPosted 4 years ago. Direct link to Moksha's post “Can somebody explain the ...” Can somebody explain the chain rule in an "easy-to-understand" way? Thanks :) • (13 votes) Seongjoo 4 years agoPosted 4 years ago. Direct link to Seongjoo's post “Consider a function (x+1)...” Consider a function (x+1)^2. 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. So to continue the example: Thus, d/dx[(x+1)^2] = 2(x+1) (37 votes) foo bar a year agoPosted a year ago. Direct link to foo bar's post “The variable power rule i...” The variable power rule is missing: • (11 votes) Sree 3 months agoPosted 3 months ago. Direct link to Sree's post “Those are just special ca...” Those are just special cases. These are the general overarching rules which apply to any old generic function f(x). (2 votes) abu 7 years agoPosted 7 years ago. Direct link to abu's post “In terms of the AP exam: ...” 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) Mark Geary 7 years agoPosted 7 years ago. Direct link to Mark Geary's post “Proofs are NOT specifical...” 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. (25 votes) Anurag 7 years agoPosted 7 years ago. Direct link to Anurag's post “In quotient rule, can't w...” In quotient rule, can't we write the derivative of f(x)/g(x) as • (2 votes) Howard Bradley 7 years agoPosted 7 years ago. Direct link to Howard Bradley's post “No, because d/dx (1/g(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) Which, with a bit of manipulation, can be made to look like the familiar quotient rule for differentiation. (9 votes) DRippy_CHexxx 10 months agoPosted 10 months ago. Direct link to DRippy_CHexxx's post “Can you apply chainrule f...” Can you apply chainrule for questions like: ddx of [f(g(h(x)))]? • (3 votes) Tanner P 10 months agoPosted 10 months ago. Direct link to Tanner P's post “Yes, you would use the ch...” Yes, you would use the chain rule twice for that case. (5 votes) darknessbright a year agoPosted a year ago. Direct link to darknessbright's post “we could also use the pow...” we could also use the power rule to prove that d/dx[x]=1 • (3 votes) Ben a year agoPosted a year ago. Direct link to Ben's post “Yes, you could, x is the ...” Yes, you could, x is the same thing as x^1, so it becomes 1x^0 when you apply the power rule, which is the same thing as 1 (5 votes) Lahlah 7 months agoPosted 7 months ago. Direct link to Lahlah's post “how do you derive the squ...” how do you derive the square root of x to the 3rd power • (1 vote) Venkata 7 months agoPosted 7 months ago. Direct link to Venkata's post “Remember that [sqrt(x)]^3...” Remember that [sqrt(x)]^3 is just x^(3/2). From here, just use the power rule. (5 votes) James a year agoPosted a year ago. Direct link to James's post “This has gotta be all for...” This has gotta be all for Calculus AB, isn't it? • (2 votes) Venkata a year agoPosted a year ago. Direct link to Venkata's post “Yes. Everything until the...” Yes. Everything until the application of integrals is AB Calc. Parametric equations and infinite sums are BC Calc (3 votes) Melia Jane 4 years agoPosted 4 years ago. Direct link to Melia Jane's post “Hey Great Learners. x(x-4...” 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). • (2 votes) Chace Caven 4 years agoPosted 4 years ago. Direct link to Chace Caven's post “For the equation `y = x(x...” For the equation Hope this help! (1 vote)Want to join the conversation?
We can see that the function has two parts, the enclosing part
(outside: ( )^2) and the enclosed part (inside: x+1).
Afterwards, you take the derivative of the inside part and multiply that with the part you found previously.
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).a^x = ln(a) a^x and differentiation rules for logarithms
f(x).g'(x)^-1 + g(x)^-1.f'(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)
Please and Thanks!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).
