Why dvexp is so different from classical math

[Filip Cernatescu]

Why the derivative of power rule is so different from the classical math?

[Mario Carneiro]

Could you be more specific? It says that the derivative of x^n is n * x^(n-1), which is pretty standard. The notation for describing what a derivative is may not be standard, but we can't easily replicate the math notation for this, which is ambiguous in several ways.

On Mon, Nov 4, 2019 at 7:25 AM 'Filip Cernatescu' via Metamath <meta...@googlegroups.com> wrote:

Why the derivative of power rule is so different from the classical math?

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[Filip Cernatescu]

Sorry! Why the dvexp proof is so different?

[Mario Carneiro]

The proof is by induction: x^1 = x so the derivative is 1 = 1 * x^(1-1), and x^(n+1) = x^n * x so the derivative is (x^n)' * x + x^n * x' = (n * x^(n-1)) * x + x^n = (n+1) * x^n.

Writing that all out in formal detail is a bit more work, but not significantly so.

Mario

On Mon, Nov 4, 2019 at 7:48 AM 'Filip Cernatescu' via Metamath <meta...@googlegroups.com> wrote:

Sorry! Why the dvexp proof is so different?

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[Filip Cernatescu]

I understand that:

The proof is by induction: x^1 = x so the derivative is 1 = 1 * x^(1-1), and x^(n+1) = x^n * x so the derivative is (x^n)' * x + x^n * x' = (n * x^(n-1)) * x + x^n = (n+1) * x^n.

but how you prove that:  (x^n)'= (n * x^(n-1)), is something recursive?

[Mario Carneiro]

It is a proof by induction; that is the inductive hypothesis.

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