Think about why the derivative of x^2 is 2x (look at binomial expansion of (x+a)^2 )
"Let epsilon > 0 be given" is a way around a vanishing but necessary variable (in this case "dx") be manipulated.
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The function name df() is not special in any way I just chose it for the happy coincidence at the end, I could have called it mashedpotatoes().
The variable name dx is not special in any way I just chose it for the happy coincidence at the end, I could have called it q or fred.
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So it is no accident that the derivative of x^n is n*x^(n-1) this falls out directly from the 2nd term binomial expansion of the (x+dx)^n
The first term is simply x^n and when the subtraction of the original x^n happens in the finite difference the first term goes away.
The second term is n*x^(n-1)*dx all other terms have dx's of higher powers. During the division of dx on both sides, this second term is the only one without any dx terms [ n*x^n-1)*dx/dx ] all the other terms have higher terms of dx, and, in the limit as dx (epsilon) goes to zero these all dissapear.
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Derivative of a sum is simply the sum of the derivatives.
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I'm using the Taylor series for Sin and Cos to find the important bits that predominate as dx epsilon goes to zero. Again I'm leaving out steps for clarity and space.
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