Question

Easier challenge:
Derivatives - Is the power rule simple?
1)Prove the power rule for integral exponents.
2)Given that result, prove that the power rule holds for all rational exponents.
3)Using any known derivative formulas/techniques, prove the power rule holds for irrational numbers as well.

Answer 1

This could take a while to write out...

Answer 2

the first one, yeah...
I'll accept shorthand :)

Answer 3

you may accept the chain rule as given...

Answer 4

Basically, if we can show it for \(x^n\), by the properties of derivatives, we'll have proved it for any polynomial \[a_0+a_1x+a_2x^2+...+a_nx^n\]So let's try and show it for \(x^n\)!

Answer 5

really? new one on me...

Answer 6

power rule can be proved via inductive method using product rule for positive integers.

Answer 7

the first one must be proved from first principles
after that accept the chain rule, and ... no more hints
besides, I wanna see different ways

Answer 8

\[\lim_{h \rightarrow 0} {f(x+h)-f(x) \over h}=\lim_{h \rightarrow 0} {(x+h)^n-x^n \over h}\]If we expand the numerator, we get\[=\binom{n}{0}x^n+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2+...+\binom{n}{n}h^n-x^n\]From this, we can cancel the \(x^n\) terms.Then, we can cancel an h from every term This leaves us with\[\lim_{h \rightarrow 0} {nx^{n-1} +h(\text{a bunch of stuff})}\]Which is equal to zero since all the "stuff" goes to 0. This leaves us with the derivative \(nx^{n-1}\).

That proves 1 from the definition of derivative without any extra assumptions.

Answer 9

This proves it for any polynomial \(a_0+a_1x+...+a_nx^n\) since we can split it apart into \(n+1\) separate polynomials and take the derivative of each, and then add them back together. We can also factor constants out of limits, so we're good there.

Answer 10

yes :)
now just assuming the chain rule and do part 2)

Answer 11

For rational exponents, we need to prove it for \[x^{1/a}\]first. Then we can simply raise this to the \(n\)-th power.

Answer 12

not quite the way I know it, so feel free to demonstrate

Answer 13

I've never done this before, so I'm winging it here.

Answer 14

you may request a hint if you like

Answer 15

\[x^r=e^{r\ln(x)}\]

Answer 16

you are doing problem 3) satellite
you can't assume to know the derivative of e^r until part 3)
at least that's the way I want to do it, but your way would work of course

Answer 17

not to be picky but if say \(\pi\) is irrational, how do we define \(x^{\pi}\) for real values of x?

Answer 18

I dunno

Answer 19

here is one i like
prove \((fg)'=f'g+g'f\) using only that \((f^2)'=2ff'\)

Answer 20

I'm gonna request a small hint for #2 here.

Answer 21

#2 use chain rule

Answer 22

\[y=x^{n/m}\implies y^m=x^n\]

Answer 23

Is implicit differentiation allowed?

Answer 24

yes, that's what I meant by chain rule

Answer 25

Well then, \[y^m=x^n \Longrightarrow my^{m-1}\;\;dy=nx^{n-1}\;\;dx\]So \[{dy \over dx}={nx^{n-1} \over my^{m-1}}\]Then, since \(y=x^{n/m}\), we know that \[\Large y^{m-1}=(n^{n/m})^{m-1}=x^{{nm \over m}-{n \over m}}=x^{n-{n \over m}}\]If we plug that into our equation for \({dy/dx}\), we have \[\Large {dy \over dx}={nx^{n-1} \over mx^{n-{n \over m}}}\]Which in turn equals \[\Large {n \over m}\cdot x^{(n-1)-(n-{n \over m})} ={n \over m} \cdot x^{{n \over m} -1}\]Which is what we wanted to prove.

Answer 26

haha, indeed! nicely done :)
(wish I could give 2 medals)
the algebra is the only pain in the butt once you know what to do

Answer 27

part 3) is actually the fastest to prove
you may use all common derivative knowledge at your disposal for that one
(chain rule, well-known derivatives and techniques, etc.)

Answer 28

Openstudy just crashed for me. Give me a minute to write it all up again.

Answer 29

I hate when that happens :(
no rush...

Answer 30

I copy what I have regularly to avoid such catastrophes

Answer 31

Using satellite's suggestion, \[\large y=x^a =e^{a \ln(x)} \Longrightarrow {dy \over dx}=e^{a \ln(x)} \cdot {a \over x}\]Simplifying this, we have \[{dy \over dx} ={ax^a \over x}=ax^{a-1}\]

Answer 32

That was an excellent exercise in the fundamentals of calculus.

Answer 33

tadah!
3/3, nice job :)
I'm glad you enjoyed it
(this is directly from the MIT calc I lectures, btw)

Answer 34

I will now post a bonus problem, which I have posted before, but did not receive a satisfactory explanation to. Let those who don't know the trick attempt it.