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I've started listening to the lectures for the MIT OpenCourseWare 18.01 Single Variable Calculus class. I understood all of it up until the teacher found the derivative of x

[tex]\frac{d}{dx} x^{n} = \frac{\Delta f}{\Delta x} = \frac{(x+\Delta x)^{n} - x^{n}}{\Delta x}[/tex]

That, I understand. Then we get to the binomial theorem to try to simplify [tex](x+\Delta x)^{n}[/tex]. The professor said that [tex](x + \Delta x)[/tex] is multiplied by itself n times, which I understand. Then he wrote:

[tex](x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + junk[/tex]

[tex]junk = O((\Delta x)^{2})[/tex] ("big O of delta x squared")

What I'm confused about is the last line. Why is that what "junk" equals? I understand that the rest of the terms don't matter as [tex]\Delta x[/tex] approaches 0, but why are they equal to what he says they are equal to "big O of delta x squared," as he said?

^{n}. Here's what he wrote on the board:[tex]\frac{d}{dx} x^{n} = \frac{\Delta f}{\Delta x} = \frac{(x+\Delta x)^{n} - x^{n}}{\Delta x}[/tex]

That, I understand. Then we get to the binomial theorem to try to simplify [tex](x+\Delta x)^{n}[/tex]. The professor said that [tex](x + \Delta x)[/tex] is multiplied by itself n times, which I understand. Then he wrote:

[tex](x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + junk[/tex]

[tex]junk = O((\Delta x)^{2})[/tex] ("big O of delta x squared")

What I'm confused about is the last line. Why is that what "junk" equals? I understand that the rest of the terms don't matter as [tex]\Delta x[/tex] approaches 0, but why are they equal to what he says they are equal to "big O of delta x squared," as he said?

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