# Mathematical Induction: Power Rule for Differentiation

1. Jul 8, 2011

### BrianMath

1. The problem statement, all variables and given/known data
Prove that
$$\frac{d}{dz}z^n = nz^{n-1}\;\;\; \forall n\in\mathbb{N}$$
using the Product Rule for differentiation and mathematical induction.

2. Relevant equations
$$\frac{d}{dz} f(z) = \lim_{\Delta z\to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z}$$
$$\frac{d}{dz}[f(z)g(z)] = f\,'(z)g(z) + g'(z)f(z)$$

3. The attempt at a solution
Let n = 1:
$$\frac{d}{dz} z = \lim_{\Delta z \to 0} \frac{z + \Delta z - z}{\Delta z} = \lim_{\Delta z \to 0} \frac{\Delta z}{\Delta z} = \lim_{\Delta z \to 0} 1 = 1$$

Assume true for n = k:
$$\frac{d}{dz} z^k = kz^{k-1}$$

Let n = k + 1:
$$\frac{d}{dz} z^{k+1} = \frac{d}{dz}[z^k\cdot z] = \frac{d}{dz} (z^k) \cdot z + \frac{d}{dz} (z) \cdot z^k = kz^{k-1}\cdot z + z^k = kz^k + z^k = (k+1)z^k = (k+1)z^{(k+1)-1}$$

Like my previous topic, I'm pretty sure I have the proof correct, but I need to make sure that it is written out correctly. I don't want to be teaching myself the wrong way to lay out proofs.

2. Jul 8, 2011

### LCKurtz

Looks fine to me.