Use the rule for differentiating a product to prove that the derivative of $x^n$ is $nx^{n-1}$ for all $n∈N$.

Question

I know the rule of differentiation, but to proving why the derivative is that is my problem. Should I be proving this question by induction because that's what I've been learning.

Answer 1

Yep! The base case should be immediate. For the inductive step, we can break off an $x$ and use Product Rule in order to apply the induction hypothesis: $$ \frac{d}{dx} \left[ x^n \right] = \frac{d}{dx} \left[ x \cdot x^{n - 1} \right] = x \cdot \underbrace{\frac{d}{dx} \left[ x^{n - 1} \right]}_{ \begin{array}{c}\text{apply ind.} \\ \text{hyp. here}\end{array}} + \frac{d}{dx} \left[ x \right] \cdot x^{n-1} = \cdots $$

Answer 2

If you assume the product differentiation rule for product of $n$ functions, you don't need induction. Simply, $\dfrac{d}{dx}(x^n) = \dfrac{d}{dx}(x\cdot x\cdot ...\cdot x) = x^{n-1}+x^{n-1}+...+x^{n-1}=nx^{n-1}.$