Question

@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.

Answer 1

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.

Answer 2

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.