# Small change (derivative)

1. Oct 5, 2009

### songoku

1. The problem statement, all variables and given/known data
Given that $$y=2+\frac{8}{x}$$. Find the approximate change in x which will cause y to increase from 6 to 6.01

2. Relevant equations
$$\frac{\delta y}{\delta x}=\frac{dy}{dx}$$

3. The attempt at a solution
I did it with two ways :

1.
when y = 6 --> x = 2
when y = 6.01 --> $$x=\frac{800}{401}$$

approximate change in x = $$\frac{800}{401}-2 = -\frac{2}{401}$$

2.
$$\delta y = 0.01$$

$$\frac{\delta y}{\delta x}=\frac{dy}{dx}$$

$$\frac{\delta y}{\delta x}=\frac{-8}{x^2}~ \text{where x = 2}$$

$$\delta x = -\frac{1}{200}$$

My question is : Can I use the first method? I don't think I can but I don't know the reason why I may not use it.

2. Oct 5, 2009

### rock.freak667

I think your first method assumes 'y' changed linearly with 'x'. But it's been a long time since I've some small changes. So it is always best to use the second method.

Also it is δy/δx ≈ dy/dx, it is only equal when you take the limit δx→0

3. Oct 5, 2009

### Dick

Your first method gives you the EXACT value of the change. The second method gives you the APPROXIMATE value of the change. 1/200=0.005. 2/401 is about 0.00499. They are nearly equal, which is what you expect.

4. Oct 5, 2009

### songoku

Hi rock.freak667 and Dick

Thanks a lot for you both !