Question

proof

Answer 1

How do we go about proving this is true

\[\frac{ d }{ dx }(x^{n}) = n(x^{n-1})\]

Answer 2

I think I seen this before. HOld on let me get my book. that's one of the differentiation proofs in real analysis

Answer 3

actually a pdf version of the book makes sense since this is a theorem and there was a proof attached to it

Answer 4

Would it be something like this?

\[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]

\[x^{n}\]

and

\[x^{n}+h\]

\[\lim_{h \rightarrow 0} \frac{ f(x+h)^{n}-f(x^{n}) }{ x^{n}}\]

Answer 5

no that's the definition of the derivative. completely different and that's another theorem

Answer 6

I thought those laws were built off of the definition of the derivative

Answer 7

Read Power rule http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx

Answer 8

damn looks like it can happen. proof 2 on the site uses it.

Answer 9

yeah, @UsukiDoll I saw the site that loser posted it involves some kind of power series.

Answer 10

wait there are two different proofs

Answer 11

from the definition?

Answer 12

there's three on that site. So I guess a proof like solving a math problem can have multiple ways to come up with the same solution as long as we're not doing anything illegal ...mathwise. ugh I'm tired. too much studying

Answer 13

technically there are infinite ways. One can always add 0

Answer 14

WHOA! no way... product rule proof is way the h*** different than in my book. I had to use sequences p_n converging to p_0 and what not to prove product rule

Answer 15

Was wondering if someone could explain

Proof 1 of the power rule in the site loser posted.

Answer 16

Just worked through the one for a constant that one is much easier.

Answer 17

these two are pretty straight forward.

\[\frac{ d }{ dx } (c) = \lim_{h \rightarrow 0} = \frac{ c-c }{ h } = \frac{ 0 }{ h }\]

\[\frac{ d }{ dx } cx = \frac{ f(cx+ch)-f(cx) }{ h } = \frac{ cx+ch-cx }{ h } = \lim_{h \rightarrow 0} = c \]