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

Question

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 $\forall n \in \mathbb{Z}_+$?

Answer 1

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.

Answer 2

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

$\frac{d}{dx}x^{n+1}=\frac{d}{dx}(x^n\times x)=nx^{n-1}\times x+x^n\times1=(n+1)x^n$

Answer 3

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 !

Answer 4

Try using the product rule.

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

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

$=x^n + x(n)x^{n-1}$

$=x^n + (n)x^n$

$=(n+1)x^{n}$

Answer 5

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, $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = \left ( k x^{k-1} \right )x + x^{k}$$

$$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = kx^{k-1+1} + x^{k}$$ $$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{k} x\right ) = kx^{k}+x^{k} $$ $$ \therefore \frac{\mathrm{d} }{\mathrm{d} x}x^{k+1}= \left ( k+1 \right )x^{k}.$$

Answer 6

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