$\frac{d}{dx}x^n=nx^{n-1}$ proof by induction

Question

Problem

Proof following statement with induction:

$$ \frac{d}{dx}x ^n=nx^{n-1} : \forall n \in \mathbb{Z}_+ $$

Attempt to solve

Base case

$$ \frac{d}{dx}x^1=1\cdot x^{0}=1 $$

which holds true

Induction step

$$ \frac{d}{dx}x\cdot x^n =_{\text{ind. hyp}} nx^{n-1}:\forall n \in \mathbb{Z}_+ $$

$$\frac{d}{dx}x^{n+1}=(n+1)x^{n}$$

It is known that product rule holds:

$$ \frac{d}{dx}f(x)g(x)=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx} $$

then

$$ \frac{d}{dx}x^{n+1}=\frac{d}{dx}x\cdot x^n =(\frac{d}{dx}x)\cdot x^n + x \cdot (\frac{d}{dx}x^n)=x^n+x(nx^{n-1})=x^n+nx^{1-1+1} $$ $$ = x^n+nx^n=(n+1)x^n \tag*{$\square$} $$

I would like to have some feedback if my solution seems correct and or not.

Answer 1

For the base case, it looks like you are doing it backwards. To illustrate:

$$\frac{d}{dx} x^1 = 1 \cdot x^0$$

Appears to apply the rule before you have shown it. Instead:

$$\frac{d}{dx} x^1 = 1 = 1 \cdot x^0$$

Also in the inductive step:

$$\frac{d}{dx} x \cdot x^n = nx^{n-1}$$

Should be:

$$\frac{d}{dx} x^n = nx^{n-1}$$

Then you can do:

$$x^{n+1} = x \cdot x^{n}$$

hence:

$$\frac{d}{dx} x^{n+1} = \frac{d}{dx} \left(x \cdot x^{n}\right)$$

and by the product rule (and so on).