# Thread: really simple multiplication prob

1. ## really simple multiplication prob

got some variables in an example that are being combined, as follows:

2x * x^2 + 4 - x^2 * 2x = 8x

does this really equal 8x?

(2x*x^2) + 4 -(x^2*2x)

both left and right cancel out leaving the 4 no?

i'm sure i'm missing something simple.

2. that lot simplifies to 4=8x

i think that was your question

3. it was more along the lines that all of the problem "2x * x^2 + 4 - x^2 * 2x" simplifying to 8x. sorry if i wasn't clear. but yeah, i'm getting 4. this is part of a derivative problem but this part doesn't involve calculus....

4. What is the exact question?

Because for you to get 8x, you need to cancel out the x^2 and have (4 * 2x) on the right...

5. Find the value of the derivative of the function $\displaystyle \frac{x^2}{x^2 + 4}$, (if it exists) at the origin.

6. Ok, so, you will need to use the quotient rule, which is if

$\displaystyle y = \frac{u}{v}$, then $\displaystyle y' = \frac{v \frac{du}{dv} - u\frac{dv}{du}}{v^2$

This gives:

$\displaystyle \frac{d(\frac{x^2}{x^2 + 4})}{dx} = \frac{(x^2 + 4).2x - x^2(2x)}{(x^2 +4)^2}$

$\displaystyle = \frac{2x^3 + 8x - 2x^3}{x^4 + 8x^2 + 16}$

Now, you get 8x.

7. EDIT: I think a bug occured as I typed my answer... ok, it seems to have been corrected. Thanks!

8. Ok, so, you will need to use the quotient rule, which is if

$\displaystyle y = \frac{u}{v}$, then $\displaystyle y' = \frac{v \frac{du}{dv} - u\frac{dv}{du}}{v^2}$

This gives:

$\displaystyle \frac{d(\frac{x^2}{x^2 + 4})}{dx} = \frac{(x^2 + 4).2x - x^2(2x)}{(x^2 +4)^2}$

$\displaystyle = \frac{2x^3 + 8x - 2x^3}{x^4 + 8x^2 + 16}$

Now, you get 8x.

9. appreciate it, that cleared it up.