Tuesday

July 22, 2014

July 22, 2014

Posted by **Ally** on Monday, October 22, 2012 at 1:01pm.

I know I take the derivative of the first one but how do i derive this one. And the second one especially don't know how to do.

- Calculus -
**Steve**, Monday, October 22, 2012 at 2:38pmUse the quotient rule to differentiate (not derive):

if y = f/g

y' = (f'g - fg')/g^2

so, we have

y = x^2/(x-2)

y' = [(2x)(x-2) - x^2(1)]/(x-2)^2

= x(x-4)/(x-2)^2

Or, you can divide first

y = x+2 + 4/(x-2)

y' = 1 - 4/(x-2)^2

anyway, min/max is where y'=0, so x=0 or 4. (This is easiest to see looking at the first form of the solution.)

which is min, which is max?

y'' = 8/(x-2)^3

y''(0) = -1, so y is concave down (max)

y''(4) = +1, so y is concave up (min)

**Related Questions**

Calculus - R=M^2(c/2-m/3) dR/dM=CM-M^2 I found the derivative. Now how would I ...

Calculus- answer check - A brand new stock is also called an initial public ...

calculus - Find the local max/min values of f using botht the first and second ...

calc. - I'm having a lot of trouble with this problem: Sketch the graph and show...

calculus - Please help with this. I submitted it below but no one responded. I ...

local min - f(x) = x^4 + ax^2 What is a if f(x) has a local minimum at x=5. How ...

calculus - (a) find the intervals on which f is incrs or decrs. (b) find the ...

Calculus - What is the point of inflection of the function f(x)=x^3-8x^2+5x+50? ...

finding numbers - The sum of two positive numbers is 20. Find the numbers if the...

Calculus - Find any absolute max/min and local max/min for the function f(x)=x^3...