## inkyvoyd @nbouscal prove the exponent rule with the product rule. Prove the product rule without the exponent rule, or any rule that requires the exponent rule. one year ago one year ago

Proof of the exponent rule with the product rule: The proof will be by induction on $$n$$. For $$n=1$$ this is simply $$f(x)=x$$, then $$f'(a)=1$$ for all numbers $$a$$, which we have already proved using the limit definition of the derivative. Now, assume that the theorem is true for $$n$$, so that if $$f(x)=x^n$$, then $$f'(a)=na^{n-1}$$ for all $$a$$. Let $$g(x)=x^{n+1}$$. If $$I(x)=x$$, the equation $$x^{n+1}=x^n\cdot x$$ can be written $$g(x)=f(x)\cdot I(x)$$ for all $$x$$; thus $$g=f\cdot I$$ It follows from the product rule that\begin{align} g'(a)=(f\cdot I)'(a)&=f'(a)\cdot I(a)+f(a)\cdot I'(a)\\ &=na^{n-1}\cdot a + a^n\cdot 1\\ &=na^n+a^n\\&=(n+1)a^n \end{align} QED. I'll prove the product rule using the limit definition in a moment.
Proof of the product rule: If $$f$$ and $$g$$ are differentiable at $$a$$, then $$f\cdot g$$ is also differentiable at $$a$$, and (f\cdot g)'(a)=f'(a)\cdot g(a)+f(a)\cdot g'(a).\\\text{ }\\ \begin{align}\text{Proof: } (f\cdot g)'(a)&=\lim_{h\to0}\frac{(f\cdot g)(a+h)-(f\cdot g)(a)}{h}\\&=\lim_{h\to0}\frac{f(a+h)g(a+h)-f(a)g(a)}{h}\\&=\lim_{h\to0}\left[\frac{f(a+h)[g(a+h)-g(a)]}{h}+\frac{[f(a+h)-f(a)]g(a)}{h} \right]\\&=\lim_{h\to0}f(a+h)\cdot \lim_{h\to0}\frac{g(a+h)-g(a)}{h}+\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\cdot \lim_{h\to0}g(a)\\ &=f(a)\cdot g'(a)+f'(a)\cdot g(a). \end{align} QED.