Let f(x)=xnf(x)=x^n. Show that f′(x)=nxn−1f'(x) = nx^{n-1}, for all x∈Rx \in \mathbb{R}.

Let n∈Nn \in \mathbb{N}. Let f:R→Rf : \mathbb{R} \rightarrow \mathbb{R}, with f(x)=xnf(x) = x^n, for all x∈Rx \in \mathbb{R}. Show that f′(x)=nxn−1f'(x) = nx^{n-1}, for all x∈Rx \in \mathbb{R}.

I was going to expand out a formal definition of derivation, and hope that the left hand side can look like the right hand side. The fact that this needs to be for all xx suggests to me that I should be thinking Induction, however I can’t find the correct setup.




Induction isn’t necessary. Have you tried using the definition? Edit: If you want to use induction, note that xn+1=x⋅xnx^{n+1}=x\cdot x^n and use the product rule.
– Git Gud
2 days ago



@GitGud If he is trying to prove something this basic, I doubt the product rule is something he is allowed to call on.
– Ken Duna
2 days ago



I tried using |f′(x)|=|f(x)−f(x+h)x+h−x||f'(x)|= |\frac{f(x) – f(x+h)}{x+h-x}| , but I normally continue this route with an analysis of the limits. Not sure how to use it here
– user381164
2 days ago


2 Answers


Start from the definition:


By definition f′(x)=limε→0(x+ε)n−xnεf'(x) = \lim_{\varepsilon \to 0} \frac{(x+\varepsilon)^n-x^n}{\varepsilon}
Since (x+ε)n−xn=nεxn−1+ε2∑nk=2(nk)εk−2xn−k(x+\varepsilon)^n-x^n= n \varepsilon x^{n-1}+ \varepsilon^2 \sum_{k=2}^n \binom{n}{k} \varepsilon^{k-2}x^{n-k}
We get f′(x)=limε→0(nxn−1+ε∑nk=2(nk)εk−2xn−k)f'(x)=\lim_{\varepsilon \to 0} (n x^{n-1}+ \varepsilon \sum_{k=2}^n \binom{n}{k} \varepsilon^{k-2}x^{n-k})
And finally f′(x)=nxn−1f'(x)=n x^{n-1}