Prove Power Rule by Math Induction and Product Rule

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
12 replies · 11K views
5hassay
Messages
81
Reaction score
0

Homework Statement



Use the Principle of Mathematical Induction and the Product Rule to prove the Power Rule when n is a positive integer.

Homework Equations



Dxxn = nxn-1

Dx(fg) = fDxg + Dxfg

The Attempt at a Solution



In summary,

Dxxn = nxn-1
Dxxk = kxk-1
Dxxk+1 = (k+1)x(k+1)-1
Dx(xkx) = (k+1)xk
xkDxx + Dxxkx = (k+1)xk
xk + kxk-1x = (k+1)xk
xk + kxk = (k+1)xk
(k+1)xk = (k+1)xk

Therefore, Dxxn = nxn-1 is valid for all positive integers n.

EDIT: Oh, and much appreciation for any help!
 
on Phys.org
Did anyone read the proof yet? If I was the professor, I might give the proof half credit. I'm not intending in any way shape or form to criticize, 5hassay. That's why students come here: to learn and improve.

On the third line of the summary, where k+1 is introduced, the proof appears to assume the result of the induction. Although for many proof methods, one can simply substitute what we need, this substitution is not allowed, because we are only allowed to assume it, the induction hypothesis, is true for k<=n.

What about arranging the proof around
xk+1= xk*x and then applying product rule to the right hand side?Also, clarify how you're going about this. Specifically, when doing induction stating the induction hypothesis, improves clarity.
For example,
Basis step: prove the statement for n=1.
... (easily forgotten/skipped, but essential)

Inductive step:
Assume for induction
Dxxk= k*xk-1

xk+1= xk*x
Dxxk+1= Dx(xk*x) Take deriv. both sides
Then apply product rule to right hand side and see what happens.
 
5hassay said:
...

The Attempt at a Solution



In summary,

Dxxn = nxn-1
This is what you assume is true. →  Dxxk = kxk-1

With the above assumption, PROVE that this is true. (Don't assume it's true.) → Dxxk+1 = (k+1)x(k+1)-1

Dx(xkx) = [STRIKE](k+1)xk[/STRIKE]
xkDxx + Dxxkx = [STRIKE](k+1)xk[/STRIKE]
xk + kxk-1x = [STRIKE](k+1)xk[/STRIKE]
xk + kxk = [STRIKE](k+1)xk[/STRIKE]
(k+1)xk = [STRIKE](k+1)xk[/STRIKE]

Write xk as x(k+1)-1
...
Pretty much what nickalh said.
 
Ahhhh, thanks Sammy, I just realized OP actually worked it out acceptably on the left hand side. You're right, ironically, deleting the right hand side and cleaning up a bit does the proof. Possibly add a concluding line, I prefer actual English instead of just notation.

Ya'll have an excellent day/afternoon/night.
 
gb7nash said:
If you really want to have a hoot, try proving the power rule from the definition of the derivative.

Haha, I do imagine that would be a good amount of fun and a good looking product afterward!
 
Last edited:
nickalh said:
Did anyone read the proof yet? If I was the professor, I might give the proof half credit. I'm not intending in any way shape or form to criticize, 5hassay. That's why students come here: to learn and improve.

On the third line of the summary, where k+1 is introduced, the proof appears to assume the result of the induction. Although for many proof methods, one can simply substitute what we need, this substitution is not allowed, because we are only allowed to assume it, the induction hypothesis, is true for k<=n.

What about arranging the proof around
xk+1= xk*x and then applying product rule to the right hand side?


Also, clarify how you're going about this. Specifically, when doing induction stating the induction hypothesis, improves clarity.
For example,
Basis step: prove the statement for n=1.
... (easily forgotten/skipped, but essential)

Inductive step:
Assume for induction
Dxxk= k*xk-1

xk+1= xk*x
Dxxk+1= Dx(xk*x) Take deriv. both sides
Then apply product rule to right hand side and see what happens.

Ah, okay. Yes, I am very new to Mathematical Induction (by definition), so I thought I would just skip the various statements. I now know that their importance is not less so as to not be involved in a summary! Also, I now understand about the n=1 step. I did indeed not think about it at all! Thanks for your help!
 
Redbelly98 said:
Looks like you have pretty much proved it. You just need to show it's true for n=1.

Okay, thanks!
 
nickalh said:
Ahhhh, thanks Sammy, I just realized OP actually worked it out acceptably on the left hand side. You're right, ironically, deleting the right hand side and cleaning up a bit does the proof. Possibly add a concluding line, I prefer actual English instead of just notation.

Ya'll have an excellent day/afternoon/night.
nickalh, I agree. I prefer English, rather than only notation, too.
 
Redbelly98 said:
Looks like you have pretty much proved it. You just need to show it's true for n=1.

5hassay said:
Okay, thanks!

You're welcome. Though what SammyS wrote in Post #7 is really correct. What you did was fine in terms of figuring things out, but a formally correct proof should follow along the lines of Post #7. Like I said, you pretty much have it.