Prove $\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$ by induction Problem
Prove $$\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$$ by mathematical induction.
 Attempt to solve 
Base case
when $n=1$
$$ \frac{d}{dx} x^1 = 1 \cdot x ^{0}=1 $$
which is true
Induction step
$$\frac{d}{dx}x^{n+1}=(n+1)x^{n+1-1}= (n+1)x^{n}$$
At this point not quite sure how to prove this with induction without proving operator $\frac{d}{dx}$ with 
$$ \frac{d}{dx}f(x)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} $$
and then proving existence of such limit with:
$$ 0<|x-a|< \delta \implies|\frac{f(a+h)-f(a)}{h}-\frac{d}{dx}f(a)| < \epsilon
$$
and then we can arrive at implication that $$\frac{d}{dx}x^n=nx^{n-1} \implies \frac{d}{dx}x^{n+1}=(n+1)x^n$$
Most likely there is easier by induction which is capable of showing that $$ \frac{d}{dx}x^n=nx^{n-1} $$ is applicable $\forall n \in \mathbb{Z}_+$?
 A: For the induction step, use the product rule.
What you already know in the induction step:


*

*You know that $\frac{d}{dx} x^n = nx^{n-1}$

*You know that $x^{n+1} = x\cdot x^n$

*You know that $\frac{d}{dx}(f(x)g(x)) = g(x)\cdot(\frac{d}{dx} f)(x) + f(x)\cdot(\frac{d}{dx}g(x))$
From that, you can calculate what $\frac{d}{dx}x^{n+1}$ should be.
A: from 1st principles by definition of the limit, if we proved that 
$f(x) = x^1$
then $$f'(x) = \frac{(x + h) - x}{h} = \frac{h}{h} = 1 = x^0$$
then 
$$f'(x^{n+1}) =  \frac{(x + h)^{n+1} - x^{n+1}}{h}$$
$$=\frac{(x + h)(x + h)^{n} - x.x^{n}}{h}$$
$$=\frac{x(x + h)^n + h(x+h)^n - x.x^n}{h}$$
$$=(x + h)^n + x\frac{(x + h)^n - x^n}{h} = A$$
now use $$f'(x^n) = \lim\limits_{h \to 0}\frac{(x + h)^n - x^n}{h} = nx^{n-1}$$ which is known for n =1
so continuing
$$A = (x + h)^n + xnx^{n-1}$$
now
$$\lim\limits_{h \to 0}(x + h)^n + xnx^{n-1}$$
$$ = x^n + n x^n$$
$$ = (n + 1) x^{n+1}$$
so given that $f'(x^n) = nx^{n+1}$
we have that 
$f'(x^{n + 1}) = (n + 1)x^{n}$
and if m = n + 1, then 
$f'(x^m) = mx^{m - 1}$ which is what we wanted to show
did you need to show it the other way too though?  i.e. going down to negative powers etc - I guess you do from the question
A: Assume that we have 
$(*) \quad \frac{d}{dx} x^n=nx^{n-1}$ 
for some $ n\in \mathbb{Z}_{+}$.
For the induction step use $(*)$ and the product rule !
A: Try using the product rule. 
$\frac{d}{dx}x^{n+1} = \frac{d}{dx}(xx^n)$
$= x^n \frac{d}{dx}x + x\frac{d}{dx}x^n$
$=x^n + x(n)x^{n-1}$
$=x^n + (n)x^n$
$=(n+1)x^{n}$
A: Base case: Let $n=1$. So $$ \text{LHS} = \frac{\mathrm{d} }{\mathrm{d} x} x^{1} = 1x^{1-1} = 1x^{0}= \text{RHS}$$
Assume by Mathematical Induction that $n=k$ for $k \in \mathbb{Z}_{+}$, i.e., $\frac{\mathrm{d} }{\mathrm{d} x}x^{k} = kx^{k-1}$. 
We want to show that the statement holds for $n=k+1$, i.e., $\frac{\mathrm{d} }{\mathrm{d} x}x^{k+1} = (k+1)x^{k+1-1}$.
Start from $$\frac{\mathrm{d} }{\mathrm{d} x}x^{k+1} = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ).$$
By Product Rule and Induction Hypothesis, $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = \left ( k x^{k-1} \right )x + x^{k}$$
$$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = kx^{k-1+1} + x^{k}$$
$$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = kx^{k}+x^{k} $$
$$ \therefore \frac{\mathrm{d} }{\mathrm{d} x}x^{k+1}= \left ( k+1 \right )x^{k}.$$
A: I ma not sure what the accepted asnwer can be, because it depends on what you are supposed to already know. 
I will propose the following.
Let's supposed this is true for $n$.
You have then $\frac{d}{dx}x^n=nx^{n-1}$
$\frac{d}{dx}x^{n+1}=\frac{d}{dx}(x^n\times x)=nx^{n-1}\times x+x^n\times1=(n+1)x^n$
