7825180
Description
Flashcards by Abigail Rejopachi, updated more than 1 year ago
|
Created by Abigail Rejopachi
about 7 years ago
|
|
| Question | Answer |
| y = ln(x) y' = 1/x | y = [ln(x^2-4)^3](2x) y' = ln(x^2-4)^3(2) + 2x[1/(x^2-4)^3]3(x^2-4)^2(2x) y' = ln(x^2-4)^3(2) + [1/(x^2-4)^3](3)(2x)(2x) y' = 2ln(x^2-4)^3 + (12x^2/x^2-4) |
| y = sen(x) y' = cos(x) y = e^x y' = e^x | y = sen(e^2x^2-1) y' = cos(e^2x^2-1)(e^2x^2-1)(4x) y' = 4xe^2x^2-1(cos e^2x^2-1) |
| y = tan(x) y' = 1/cos^2(x) | y = tan (x^2-1) y' = [1/cos^2(x^2-1)](2x) y' = 2x/cos^2(x^2-1) |
| y = ln(x) y' = 1/x | y = ln (x^2 - 2x+1) y' = 1/x^2 - 2x+1 y' = [1/x^2 - 2x+1](2x-2) y' = 2x-2/x^2 - 2x+1 |
| y = cos(x) y' = -sen(x) | y = cos[ln(x)] y' = -sen[ln(x)](1/x) y' = -sen[ln(x)]/x |
Want to create your own Flashcards for free with GoConqr? Learn more.