Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
@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
@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

This Question is Closed

nbouscalBest ResponseYou've already chosen the best response.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 \(f'(a)=na^{n1}\) 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.
 one year ago

nbouscalBest ResponseYou've already chosen the best response.1
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.
 one year ago
See more questions >>>
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.