You are here:

Calculus/derivatives

Advertisement


Question
I have this problem and I've found the first derivative:
f(x)= x^3/3-x
f'(x)=(1/3x)(3x-1)
x= +-1
I don't know how to acquire the second derivative. Could you please show all work and explain.  Thank you very much

Answer
I'm not sure where f'(x) = (1/3x)(3x-1) came from, but it can be rewritten as (3x-1)/(3x).

If the function is f(x) = x³/(3-x), then this is f(x) = g(x)/h(x), where g=x³ and h=3-x.
The definition of the derivative is f'(x) = [h(x)g'(x) - g(x)h'(x)]/h²(x).  

Since it can be seen that g'(x) = 3x² and h'(x) = -1, this gives
f'(x) = ((3-x)3x² - x³(-1))/(3-x)². This reduces to
f'(x) = (9x² - 3x³ + x³)/(3-x)² = (9x² - 2x³)/(3-x)².

Given that f'(x) = (9x² - 2x³)/(3-x)², we would then take g(x) = 9x² - 2x³ and h(x) = (3-x)².
To find f"(x), it is (h(x)g'(x) - g(x)h'(x))/h²(x).  It can be seen here that g'(x) = 19x - 6x² and h'(x) = -2(3-x).

So plug g(x) =  9x² - 2x³, h(x) = (3-x)², g'(x) = 19x - 6x², and h'(x) = 2x-6 into
f" = (hg' - gh')/h² to get the answer.

Calculus

All Answers


Answers by Expert:


Ask Experts

Volunteer


Scotto

Expertise

Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

Experience

Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

Publications
Maybe not a publication, but I have respond to well oveer 7,500 questions on the PC. Well over 2,000 of them have been in calculus.

Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

Awards and Honors
I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

Past/Present Clients
My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.

©2012 About.com, a part of The New York Times Company. All rights reserved.