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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?

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That’s correct. –  Adeel Feb 8 '12 at 1:47
    
There's a problem with the base case. How do we know that the derivative of $z$ is $1$? –  Patrick Mar 11 '12 at 0:37

2 Answers 2

up vote 8 down vote accepted

You have all the right work, but you need to polish it up a bit.

For the base case, show sequentially how you start with the base and end up with what the claim shows:

$$\frac{\mathrm{d}}{\mathrm{d}x} z^1 = 1 = 1 \cdot z^{1 - 1}$$

So the claim holds for the base case. (It actually holds for $n = 0$ as well, so verify what you want your base to be.)

Next, assume that the claim holds for all $k \geq 1$. Then if we can show that that assumption implies that the claim holds for $k + 1$, then the claim holds for all $n \geq 1.$

So, then we have

$$ \frac{\mathrm{d}}{\mathrm{d}x} z^{k+1} = \frac{\mathrm{d}}{\mathrm{d}x} z^k \cdot z = z \cdot \frac{\mathrm{d}}{\mathrm{d}x} z^k + z^k \cdot \frac{\mathrm{d}}{\mathrm{d}x} z = z \cdot (kz^{k-1}) + z^k = kz^k + z^k = (k+1)z^k$$

Thus the claim holds for all $n \geq 1$.

This is something to really pay attention to when proving anything. It is subtle but makes proofs much more readable. Start with what you know, and continue until you get to the conclusion that you want. You don't want to start with the thing that you want and manipulate until you get what you have. While in many situations it is equivalent in validity, it is not always and is never the best technique for showing a claim is true.

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Ok thats cool, ty –  Jim_CS Feb 8 '12 at 2:13

The induction step is obvious if one works with logarithmic derivatives.

$\qquad\ (fg)' =\: f\:'g + fg'\ $ when divided through by $fg$ becomes

$\qquad\dfrac{(fg)'}{fg}\: =\: \dfrac{f\:'}f + \dfrac{g'}g\ $ which, in terms of $\:D(f) := \dfrac{f\:'}f\: $ becomes

$\qquad D(fg)\: =\: D(f) + D(g)\ $ the "logarithm-like" law of the logarithmic derivative $D$

$\ \Rightarrow\ D(x^n)\: =\: n\: D(x)\ $ by an obvious induction (prove it!)

$\ \Rightarrow\ \dfrac{(x^n)'}{x^n}\: =\: n\:\dfrac{1}x $

$\ \Rightarrow\ \: (x^n)' =\: n\ x^{n-1}$

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2  
I'm so glad that this is obvious for you. –  The Chaz 2.0 Feb 8 '12 at 4:17
2  
Elaborating on what @TheChaz has said, I think the word obvious is dangerous in helping someone with a mathematical concept, especially a formal proof. That which is obvious for one person is not immediately obvious for everyone. –  Kurtis Zimmerman Feb 8 '12 at 4:32
    
@Kurtis The point is that many inductive proofs can be greatly simplified by transforming the problem in a way that makes the inductive step so simple that the inductive proof becomes "obvious" (e.g. that $1^n = 1$). For many further examples see here. –  Math Gems Feb 8 '12 at 7:07
    
@TheChaz I'd be happy to elaborate if you could please clarify what you mean by that comment. –  Math Gems Feb 8 '12 at 7:09
    
I meant what @Kurtis said. While I appreciate a "slick" approach to problems, the implication here seems to be that the student is restricted to the product rule. Maybe he should just write "the proof is trivially obvious" on the page and submit that... :) –  The Chaz 2.0 Feb 8 '12 at 7:39

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