Solving for x: Setting dA/dx = 0

[musicmar]

Homework Statement

I need to solve for x by setting the derivative of A equal to 0.
A=(x^2)/2(x-2)

The Attempt at a Solution

dA/dx= x(x-2)^-1(-1)((1/2)x^2)

= (-x^3)/2(x-2)

If someone could check this so far and point me in the right direction, that would be great.

 

[arildno]

Learn the rules of differentiation, this is just nonsense.

 

[lurflurf]

does
A=(x^2)/2(x-2)
mean
A=(x^2)/[2(x-2)]
?
If so
A=(x^2)/[2(x-2)]=A=(x^2-2)/[2(x-2)]+2/[2(x-2)]=(x+2)/2+1/(x-2))

 

[musicmar]

Sorry, I've looked back and realize that it was really confusing.
Here is a more comprehensible restatement.

A=(x^2)/[2(x-2)]

Is the derivative of A (-x^3)/[2(x-2)],
or did I do something completely wrong?

 

[arildno]

Yes, it is utterly wrong.

A(x) is a fraction!

We may write it as:

$$A(x)=\frac{f(x)}{g(x)}$$

1. Exercise: What is f(x)?
2. Exercise: What is g(x)?

Furthermore, the derivative of a fraction is given as:
$$A'(x)=\frac{f'(x)*g(x)-f(x)*g'(x)}{g(x)^{2}}(*)$$

3. Exercise: What is f'(x)?
4. Exercise: What is g'(x)?

5. Put the correct expressions into (*), and simplify the expression for A'(x)!

 

[lurflurf]

hint simplify A
A=(x^2)/[2(x-2)]=(x^2-2)/[2(x-2)]+2/[2(x-2)]=(x+2)/2+1/(x-2)