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## inkyvoyd 3 years ago @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.

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1. nbouscal

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.

2. nbouscal

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.

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