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Mathematics 7 Online
OpenStudy (ahsome):

What does this equation mean?

OpenStudy (ahsome):

\[f`(x)=\lim_{h \to 0}\frac{f(x_0+h)-f(x_0)}{h}\] Or: \[f`(x)=\lim_{\triangle x \to 0}\frac{f(x_0+\triangle x)-f(x_0)}{\triangle x}\]

OpenStudy (ahsome):

We are learning the basics of calculus, and this came up

OpenStudy (jtvatsim):

This is the definition of the derivative. It tells you how to find the slope of the line tangent to the curve f(x) at the point x.

OpenStudy (jtvatsim):

There is much more to it than that however, but that's the basic idea. :)

OpenStudy (jtvatsim):

Check out this video: https://www.youtube.com/watch?v=VOIUtvAdIgs

OpenStudy (ahsome):

So, its telling me how to find the slope of a line that hits the curve \(f(x)\) once, at the point \(x\)?

OpenStudy (jtvatsim):

Yes, that is correct!

OpenStudy (jtvatsim):

You can think of the derivative as the "slope of a curve"

OpenStudy (ahsome):

So kinda like this?|dw:1417572768469:dw|

OpenStudy (jtvatsim):

It's based on the same ideas that you used in algebra to find the slope of a line.

OpenStudy (ahsome):

So it gives the slope for that line?

OpenStudy (jtvatsim):

That is close... the long equation is precisely the slope of that line, exactly

OpenStudy (jtvatsim):

Note that it doesn't give you the equation of the ENTIRE line, just the SLOPE

OpenStudy (jtvatsim):

That video I posted explains it in more detail.

OpenStudy (jtvatsim):

It's only about 3 minutes

OpenStudy (ahsome):

Oh, K

OpenStudy (ahsome):

Yup. But then, since you also know the point where it hits \(f(x)\), couldn't you just sub those values into: \(y-y_1=m(x-x_1)\)?

OpenStudy (ahsome):

To find the equation for the line?

OpenStudy (jtvatsim):

Of course, that is considered a more advanced technique called "Linearization" basically figuring out the line that matches the curve at a given point. You've got the idea!

OpenStudy (ahsome):

Ok, thanks so much @jtvatsim :D

OpenStudy (jtvatsim):

No problem, and good luck with your studies! :)

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