A community for students.
Here's the question you clicked on:
 0 viewing
inkyvoyd
 4 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.
inkyvoyd
 4 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.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Proof 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 \(f'(a)=na^{n1}\) 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\[\begin{align} g'(a)=(f\cdot I)'(a)&=f'(a)\cdot I(a)+f(a)\cdot I'(a)\\ &=na^{n1}\cdot a + a^n\cdot 1\\ &=na^n+a^n\\&=(n+1)a^n \end{align}\] QED. I'll prove the product rule using the limit definition in a moment.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Proof 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.