|
Unlimited Tutoring & Homework Help
|
|
Long chain rule
Longest chain consisting of six carbon atoms..
Longest chain consisting of six carbon atoms..Naming the side chains (or substituent)
If there is any other substituent on the chain, the compound is named as the derivative of the corresponding alkene. The side chains (or substituents) are named as per rules for alkane..
Chain Rule for Finding Derivatives - Two quick and basic examples! For more free math videos, visit JustMathTutoring.com - I have organized links to over 100 FREE math videos made by me!
Derivatives - Product + Chain Rule + Factoring - A quick example for a friend out there in internet land! For more free math videos, check out JustMathTutoring.com
Question : Hi, everyone. I need to find the derivative of the following problem using the chain rule:
[1 + sin^3(x^5)]^12
Thank you so much!
Answer : Apply the chain rule four times: d/dx[1 + sin^3(x^5)]^12 = 12*[1 + sin^3(x^5)]^11 * d/dx[1 + sin^3(x^5)] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * d/dx[sin(x^5)] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * cos(x^5) * d/dx[x^5] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * cos(x^5) * 5x^4.. More from Yahoo Answers
Answer : Apply the chain rule four times: d/dx[1 + sin^3(x^5)]^12 = 12*[1 + sin^3(x^5)]^11 * d/dx[1 + sin^3(x^5)] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * d/dx[sin(x^5)] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * cos(x^5) * d/dx[x^5] = 12*[1 + sin^3(x^5)]^11 * 3sin^2(x^5) * cos(x^5) * 5x^4.. More from Yahoo Answers
Question : I need to use the derivative to find a different answer but I cant figure out the derivative.
What is the derivative of x/(sqrt(2x-1))
I am pretty sure we are supposed to use the chain rule.
Answer : x/ (2x-1) we are going to use this formula u/v = (v d/dx u - u d/dx v ) / v or the same as ( g f ' - f g' ) g u = x v = (2x-1) [ (2x-1) d/dx x - x d/dx (2x-1) ] / (2x-1) [ (2x-1)^1/2 (1) - (x) (1/2 (2x-1)^-1/2 (2)) ] / 2x-1 (2x-1) - x(2x-1)^-1/2 / 2x-1.. More from Yahoo Answers
Answer : x/ (2x-1) we are going to use this formula u/v = (v d/dx u - u d/dx v ) / v or the same as ( g f ' - f g' ) g u = x v = (2x-1) [ (2x-1) d/dx x - x d/dx (2x-1) ] / (2x-1) [ (2x-1)^1/2 (1) - (x) (1/2 (2x-1)^-1/2 (2)) ] / 2x-1 (2x-1) - x(2x-1)^-1/2 / 2x-1.. More from Yahoo Answers