3 ms·
Ooh, that's a neat technique. I think you can get an expression for the derivative from it. Say we want the derivative of f(x) = x^3 - 2x^2 + 5. That becomes:
by justinpombrio 6y ago
Ooh, that's a neat technique. I think you can get an expression for the derivative from it.
Say we want the derivative of f(x) = x^3 - 2x^2 + 5. That becomes:
(x+e)^3 - 2(x+e)^2 + 5
= x^3 + 3x^2e - 2x^2 -4xe + 5
= (x^3 - 2x^2 + 5) + (3x^2 - 4x)e
The term in front of 'e' is "3x^2 - 4x", which is f'(x).
- judofyr 6y agoJust to clarify: This result follows directly from the definition of the derivative: f'(x) = lim_{e->0} (f(x+e) - f(x))/e If you're able to express f(x+e) on the form (f(x) + y e) then it follows that y is the derivative. It also should be noted that auto-diff doesn't let you skip the rules for derivation and you're using the same calculation as you would to show that e.g. f'(x^n)=n*x^{n-1}.