Proof for y=x^n

[rbcobra]

I have to set up a proof for y=x^n showing how its derivative is dy/dx=nx^(n-1). I am given 6 parts of the proof that i just need to arrange into the correct order. They are,

a) (x+h)^n = x^n + nx^(n-1)h + (n/2)(n-1)x^(n-2)h^2 +...h^n
b) (x+h)^n - x^n = nx^(n-1)h + (n/2)(n-1)x^(n-2)h^2 +...h^n
c) lim[sub:3njnb1xy]h-> 0[/sub:3njnb1xy] [(x+h)^n - x^n]/h = nx^(n-1)
d) dy/dx = lim[sub:3njnb1xy]h-> 0[/sub:3njnb1xy] [(x+h)^n - x^n]/h
e) y=x^n
f) [(x+h)^n - x^n]/h = nx^(n-1) + h[n/2(n-1)x^(n-2) +... h^(n-2)

So my thoughts were e,d,f,b,a,c. I think that e,d should be first... and b comes before a...However I was hoping I could get some feedback, so I would know if I am even close. I am not use to using limits so this question has really thrown me. Thanks in advance!
RC

 

[Deleted member 4993]

I have to set up a proof for y=x^n showing how its derivative is dy/dx=nx^(n-1). I am given 6 parts of the proof that i just need to arrange into the correct order. They are,

a) (x+h)^n = x^n + nx^(n-1)h + (n/2)(n-1)x^(n-2)h^2 +...h^n
b) (x+h)^n - x^n = nx^(n-1)h + (n/2)(n-1)x^(n-2)h^2 +...h^n
c) lim[subx5l0e4n]h-> 0[/subx5l0e4n] [(x+h)^n - x^n]/h = nx^(n-1)
d) dy/dx = lim[subx5l0e4n]h-> 0[/subx5l0e4n] [(x+h)^n - x^n]/h
e) y=x^n
f) [(x+h)^n - x^n]/h = nx^(n-1) + h[n/2(n-1)x^(n-2) +... h^(n-2)

So my thoughts were e,d,f,b,a,c. I think that e,d should be first

I would put 'a' after e,d

... and b comes before a...However I was hoping I could get some feedback, so I would know if I am even close. I am not use to using limits so this question has really thrown me. Thanks in advance!
RC

 

[galactus]

Let \(\displaystyle f(x)=x^{n}\)

Use the definition of a limit:

\(\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{x\to 0}\frac{(x+h)^{n}-x^{n}}{h}\)

Expand \(\displaystyle (x+h)^{n}\) by using the binomial theorem:

This is kinda messy:

\(\displaystyle f'(x)=\lim_{h\to 0}\frac{\left[x^{n}+nx^{n-1}h+\frac{n(n-1)}{2!}x^{n-2}h^{2}+......+nxh^{n-1}+h^{n}\right]-x^{n}}{h}\)

Cancel a factor of h:

\(\displaystyle \lim_{h\to 0}\left[nx^{n-1}+\frac{n(n-1)}{2!}x^{n-2}h+.....+nxh^{n-2}+h^{n-1}\right]\)

Every term but the first has a factor of h, so every term but the first approaches 0 as h-->0.

So, we get \(\displaystyle f'(x)=nx^{n-1}\)

 

[rbcobra]

ok, so since the final terms all have the h except for nx^(n-1) in (f) that would put that last... but i am pretty sure that (a) has to come after (b) because of the subtraction of x^n from the left side and addition to the right side. And (c) is the limit without the binomial expansion added yet, so that needs to come before (b) and (a). Wow my brain... okay so now i have changed it to e,d,c,b,a,f. Does that seem possible or is my logic off? This is the worst kind of jigsaw puzzle.

 

[dts5044]

in e you are just stating the equation, so obviously e is first

in d you are stating the definition of the derivative, so d is second

(c) is not the limit without the binomial step, (c) is the conclusion to the problem!

because the derivative of x^n is n * x^(n-1)

and (c) states: limh-> 0 [(x+h)^n - x^n]/h = nx^(n-1)

so on the left side of the equation you have the definition of the derivative applied to y = x^n, and on the right side of the equation you have the solution you are ultimately looking for, so c should come last

galactus' post writes up the solution very nicely. knowing that e and d come first and c goes last, try and put the rest in order by looking at his post. He has explained it very clearly.

*edit: if you are confused about it, you should note that the derivative of a function y can be expressed in different ways:

derivative of y = dy/dx = lim (as h goes to 0) [f(x+h) - f(x)] / h

they are all different ways of saying the same thing