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Mathematics 13 Online
OpenStudy (anonymous):

Explain why (see picture) below

OpenStudy (anonymous):

OpenStudy (anonymous):

if you have you text book, look at the section that says 'derivatives' it should be some where around chapter 4 in you calculus book. while youre doing that i'll try to answer too!

OpenStudy (anonymous):

Unfortunately I don't have textbook, I've been trying to learn calc online. Thanks!

OpenStudy (anonymous):

recall the equation for slopes is: \[m = \frac{ y_2 - y_1 }{ x_2 - x_1}\] and remember how f(x) is a fancy way of saying y?

OpenStudy (anonymous):

so the slope equation now looks like this: \[m = \frac{ f(x) - f(x) }{ x_2 - x_1}\]

OpenStudy (anonymous):

I understand what a derivative is, what I don't under stand is (see attachment)

OpenStudy (anonymous):

now lets make... \[x_2 - x_1 = h\] so the slope equation now looks like: \[m = \frac{ f(x) - f(x) }{ h}\] we will now have the first f(x), and have x = x+h so

OpenStudy (anonymous):

good to know, i can't ever tell where people are exactly at when they ask questions. so if you know what a derivative is, then looking at the equation, its saying u'(x) - u'(x) = 0

OpenStudy (anonymous):

Sorry for the confusion. But why does the limit of u'(x) as h->0 equal u'(x)?

OpenStudy (anonymous):

'the limit of u'(x) as h->0' is a formal way of saying u'(x)

OpenStudy (anonymous):

Everything's clear now, really needed that. Thanks!

OpenStudy (anonymous):

you're welcome, sorry i didn't get to your actual question sooner!

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