Explain Differential

[skatru]

I'm not exactly sure how to go about finding differentials. Could someone explain them to me?

This is what I know but I don't understand it.

dy = f'(x)dx
change in y = f'(x)(change in x)

(2.001)[sup:1wxeo0cr]5[/sup:1wxeo0cr]

Using differentials it equals 32.08 but how do you figure this out.

Thank you.

 

[stapel]

I'm not exactly sure how to go about finding differentials.

So you're saying that you don't know how to differentiate, how to find derivatives? Or just that you don't understand how to use the plug-and-chug formula for using differentials to find linear approximations?

If the former, then you need to hire a qualified local tutor and spend many hours a week in concentrated re-learning of the past few weeks or months of material. If the latter, please reply showing where in the plug-and-chug formulaic process you are getting stuck.

For instance, for 2.001[sup:3hnj5c82]5[/sup:3hnj5c82], obviously you would be working with f(x) = y = x[sup:3hnj5c82]5[/sup:3hnj5c82], x = 2, and dx = 0.001. So you differentiated, plugged the various values into the formula, and... then what?

Please be complete. Thank you!

Eliz.

 

[PAULK]

I'm not exactly sure how to go about finding differentials. Could someone explain them to me?

This is what I know but I don't understand it.

dy = f'(x)dx
change in y = f'(x)(change in x)

(2.001)[sup:2f359y6m]5[/sup:2f359y6m]

Using differentials it equals 32.08 but how do you figure this out.

Thank you.

When you write:

change in y = f-prime(x)(change in x)

[Sorry, but when I type f', I can't see the 'prime' sign AT ALL using this awful typeface.]

You really mean:
change in y is approximately f-prime(some convenient x)(change in x)

So to APPROXIMATE (2.001)^5 [not to compute it exactly -- that requires a very good calculator.] here is what you do:

1. You write y = f(x) = x^5 as an approximating function.
2. You take x = 2 as 'some convenient x' and use it to compute f(2) AND fprime(2). You need them both.
3. You take 0.001 as 'change in x'
4. You compute a 'change in y' by the approximation above:

Now fprime(x) = 5x^4.
fprime(2) = 80
change in y = 80(0.001) = 0.08.
y = f(2) + change in y = 32 + 0.08

Note that 2.001^5 IS NOT EQUAL TO 32.08, just approximately.

 

[skatru]

Sorry about not being more clear. I know how to differentiate.

I can answer F'(x)(change in x) I know that if I use 2 it will equal .08. I don't get the change in y part.

Looking at what PaulK wrote, which is very helpful, I don't get why y = f(x) + change in y.

I'm not sure why that happens or where it comes from. That's the part that really confuses me.

Thanks.

 

[stapel]

...I don't get the change in y part....I don't get why y = f(x) + change in y. I'm not sure why that happens or where it comes from. That's the part that really confuses me.

The lessons in the link (provided earlier) should go a long way toward explaining the origin and "sense" of the formula. :idea:

Eliz.

 

[skatru]

Thanks, I didn't notice there was a link but I see it now.