Question

Find the gradient of f(x)=ax^n by using first principles.

Answer 1

there should be some factorization formula ....

\(a^n - b^n = (a-b)(some other term) \)

http://forums.futura-sciences.com/mathematiques-college-lycee/97289-factorisation-de-a-n-b-n.html

Answer 2

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

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

using binomial expansion

\[a \times \lim_{h \rightarrow 0}\frac{[\left(\begin{matrix}n \\ 0\end{matrix}\right)x^n + \left(\begin{matrix}n \\ 1\end{matrix}\right) x^{n-1}h +... + h^n] - x^n}{h}\]

Answer 3

now nC0 is always 1

so the x^n terms eliminate on the top

Answer 4

now dividing by h leaves

nC1 x^(n-1) + h( the rest of the expansion)

letting h -> 0

we get nx^(n-1)

Answer 5

\[\lim_{h\to0}{a^{x+h}-a^x\over h}=a^x\lim_{h\to0}{a^h-1\over h}\]oh I thought we were doing this one..
Hey @eigenschmeigen we did this yesterday!

Answer 6

i missed out how nC1 is always n

Answer 7

waa whats going on im confused

Answer 8

I was doing the wrong prob
but wasn't it you or.... oh no, wait it was callisto who did this same problem with me yesterday
sorry, nevermind :P

Answer 9

hahaha i am so confused now

Answer 10

ok i think i understand

Answer 11

did this prob yesterday
thought it was with you
must have been someone else
that's all :)

Answer 12

for the a^x i think we would need to write it as e^(xlna)

Answer 13

@TuringTest misread question

Answer 14

@eigenschmeigen thats right ... change a to e

Answer 15

yes, I keep saying that^
but anyway, blathering on\[\lim_{h\to0}{a^{x+h}-a^x\over h}=a^x\lim_{h\to0}{a^h-1\over h}\]so now we gotta prove that\[\lim_{h\to0}{a^h-1\over h}=\ln a\]which I think is usually done graphically if we pretend not to know the derivative of e^x

Answer 16

yeah:

\[\lim_{h \rightarrow 0} \frac{a^{x+h} - a^x}{h} =\lim_{h \rightarrow 0} \frac{e^{lna(x+h)} - e^{xlna}}{h}\]

now factor out e^xlna

multiply inside limit by lna/lna

let s = hlna
as h-> 0 , s-> 0
we get lna(e^s - 1)/s

Answer 17

ah ok

Answer 18

yeah that works, but it requires you to know the derivative of e^x, which we assume they don't
MIT has a good (but long) derivation based on graphs
yours is perfectly valid though :)

Answer 19

ooh i'd like to see - is there a video or transcript available online? i know they put some stuff online

Answer 20

here it is:
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/session-17-the-exponential-function-its-derivative-and-its-inverse/
yeah it's cool, just follow closely

Answer 21

thanks :)

Answer 22

no prob, enjoy !