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-Hackensack-

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-Hackensack-

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AKG

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Hmm... You mean from first principles?hackensack said:

-Hackensack-

In case you don't already have it, there is a general rule for functions of the form:

[tex]f(x)\ =\ a^{u(x)}[/tex]. That rule is:

[tex]f'(x)\ =\ (a^{u(x)})[\ln(a)][u'(x)][/tex].

I'll have to think about how one would go about doing this from first principles. See http://library.thinkquest.org/C0110248/calculus/difnexp.htm?tqskip1=1&tqtime=0827 for proof of the exponential derivative rule.

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mathman

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The derivative of the r.h.s is:

ln2.e

To get the same result from first principles, expand the r.h.s. of the first equation in a power series, with x replaced by a (a=deltax), when you work on the difference quotient:

(2

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Zurtex

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[tex]y = a^x[/tex]

[tex]\ln y = \ln \left( a^x \right)[/tex]

[tex]\ln y = x \ln (a)[/tex]

[tex]\frac{dy}{dx} \left( \frac{1}{y} \right) = \ln (a)[/tex]

[tex]\frac{dy}{dx} = y \ln(a)[/tex]

[tex]\frac{dy}{dx} = a^x \ln(a)[/tex]

Hope that helps, whenever differentiating [itex]f(x)^{g(x)}[/itex] it can be tackled in the same way.

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HallsofIvy

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If youhackensack said:

-Hackensack-

y(x+h)- y(x)= 2

(y(x+h)- y(x))/h= 2

The derivative is the limit of that as h->0.

Notice that the "2

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(2^x) Lim ((1+1)^h -1)/h =

h-->0

=(2^x)Lim(1+h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - +(-1))/h=

h-->0

=(2^x)Lim(h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - -)/h =

h-->0

=(2^x)Lim(1+(h-1)/2!+(h-1)(h-2)/3!+ - - - -)

h-->0

=(2^x)(1-1/2!+2!/3!-3!/4!+ - + - + -)= (2^x)(1-1/2+1/3-1/4+1/5- + - +)

=(2^x)(ln2). I've just finished HallsofIvy's work.

h-->0

=(2^x)Lim(1+h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - +(-1))/h=

h-->0

=(2^x)Lim(h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - -)/h =

h-->0

=(2^x)Lim(1+(h-1)/2!+(h-1)(h-2)/3!+ - - - -)

h-->0

=(2^x)(1-1/2!+2!/3!-3!/4!+ - + - + -)= (2^x)(1-1/2+1/3-1/4+1/5- + - +)

=(2^x)(ln2). I've just finished HallsofIvy's work.

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implicit diffy.

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cookiemonster

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HallsofIvy

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If you were referring to wisky40's work, that was just an expansion using the binomial theorem, known long before calculus itself (and regularly used to find derivatives in the early days of calculus).cookiemonster said:

cookiemonster

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Gokul43201

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Tom Mattson

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What is the statement P(n)? Is it this:Gokul43201 said:

If so, then the proof (if it can be done inductively) will only be good for integer bases, and therefore won't apply to, say, base e. Zurtex's post, on the other hand, covers all real bases.

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The binomial expansion is simply the taylor series of (x + 1)^n, is it not?HallsofIvy said:If you were referring to wisky40's work, that was just an expansion using the binomial theorem, known long before calculus itself (and regularly used to find derivatives in the early days of calculus).

cookiemonster

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HallsofIvy

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Therefore what? A power series expansion ofcookiemonster said:The binomial expansion is simply the taylor series of (x + 1)^n, is it not?

cookiemonster

For example, it is easy to see that the Taylor's series (strictly speaking "McLaurin series") for f(x)= 1/(1- x) is 1+ x+ x

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Can somebody explain how to find the n'th derivitive of f(x)=2^x. After getting the first derivitive finding the second derivitive is giving me alot of trouble.

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Gokul43201

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since ln2 is a constant, independent of x.

This can be repeated any number of times to give :

[tex] \frac {d^n} {dx^n} 2^x = (ln2)^n 2^x [/tex]

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Zurtex said:[tex]\frac{dy}{dx} \left( \frac{1}{y} \right) = \ln (a)[/tex]

[tex]\frac{dy}{dx} = y \ln(a)[/tex]

you can do that?

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AKG

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Perhaps you misunderstood what Zurtex wrote:

(dy/dx) * (1/y) = ln(a)

dy/dx = y*ln(a)

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let y=2 ^ x

then log y = log 2^x

log y= x . log2

now, u can easily do this

then log y = log 2^x

log y= x . log2

now, u can easily do this

- #19

mathwonk

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I posted a complete answer to this question today in the thead "introducing logarithms"

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