Here's a question:
Derive $\frac{\mathrm{d}}{\mathrm{d}z} z^n = nz^{n-1},$ when n is a positive integer by using mathematical induction and and the derivative of a product of two functions \frac{\mathrm{d}}{\mathrm{d}z} f(z)g(z) = f(z)g'(z) + f'(z)g(z).
Here's what I did. Is this mathematical induction performed correctly?
Base case: n = 1 $\frac{\mathrm{d}}{\mathrm{d}z} z^1 = 1z^{1-1}$ $1 = 1$
Base case is true.
Inductive step: Assume true for $n = k$, show true for $k + 1$.
So show: $\frac{\mathrm{d}}{\mathrm{d}z} z^{(k+1)} = (k+1)z^{(k+1) - 1} = (k+1)z^k$
$\frac{\mathrm{d}}{\mathrm{d}z} z^{(k+1)} = \frac{\mathrm{d}}{\mathrm{d}z} z^kz.$
Using derivative of a product of two functions: $\frac{\mathrm{d}}{\mathrm{d}z} z^kz =z^k\cdot 1 + kz^{k-1}\cdot z$ $ =z^k + kz^k$ $ = (k+1)z^k$
So as the base case holds and the inductive step holds, this means the original statement is valid?