Proof by induction (power rule of the derivative)

Question

Using the differentiation formulas $\displaystyle\frac{d}{dx}x=1$ and $\displaystyle\frac{d}{dx}(fg)=f\frac{dg}{dx}+g \frac{df}{dx}$, prove that $$\frac{d}{dx} x^n=nx^{n-1}$$ for all natural number $n$. Thanks!

Answer 1

The base case is obvious. suppose $ (x^n)' = nx^{n-1} $, we must show that $(x^{n+1})' = (n+1)x^n $. Notice

$$ (x^{n+1})' = (x^n \cdot x )' = (x^n)' \cdot x + x^n \cdot 1 =_{(*)} (nx^{n-1})x + x^n = nx^n + x^n = (n+1)x^n $$

And the result holds by mathematical induction. $(*)$ holds by the induction hyphotesis.