derivitive of 2^x

[hackensack]

Hey, ive been tiring over this problem for a while....how do you get the derivitive of y=2^x using the definition of derivitive? Any help would be greatly appreciated.

-Hackensack-

[AKG]

Hey, ive been tiring over this problem for a while....how do you get the derivitive of y=2^x using the definition of derivitive? Any help would be greatly appreciated.

-Hackensack-Hmm... You mean from first principles?

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

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

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

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

[mathman]

The simplest way to get the derivative is to use the equation:
2x=exln2
The derivative of the r.h.s is:
ln2.exln2=ln2.2x

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:
(2x+a-2x)/a

[Zurtex]

I realise this has been posted up two times but they actually both confused me so I'd like to see if I can help make it a little clearer. (a is a constant)

y = a^x

\ln y = \ln \left( a^x \right)

\ln y = x \ln (a)

\frac{dy}{dx} \left( \frac{1}{y} \right) = \ln (a)

\frac{dy}{dx} = y \ln(a)

\frac{dy}{dx} = a^x \ln(a)


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

[HallsofIvy]

Hey, ive been tiring over this problem for a while....how do you get the derivitive of y=2^x using the definition of derivitive? Any help would be greatly appreciated.

-Hackensack-

If you really mean "using the definitionof the derivative then you would start this way:

y(x+h)- y(x)= 2x+h- 2x= 2x(2h-1) so

(y(x+h)- y(x))/h= 2x((2h- 1)/h)

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

Notice that the "2x" factor does not involve h while the "(2h-1)/h" factor does not involve x. The limit will be C2x where the constant C is the limit of (2h-1)/h. The hard part is proving that that limit is ln(2).

[wisky40]

:wink: (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.

[Ebolamonk3y]

implicit diffy.

[cookiemonster]

I imagine that using a series expansion would involve the derivative anyway. May as well use L'Hospital's if you can do that...

cookiemonster

[HallsofIvy]

I imagine that using a series expansion would involve the derivative anyway. May as well use L'Hospital's if you can do that...

cookiemonster

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).

[Gokul43201]

I haven't tried this but i imagine this {lim(2^h - 1)/h = ln(2), as h -->0} can also be proven inductively.

[Tom Mattson]

I haven't tried this but i imagine this {lim(2^h - 1)/h = ln(2), as h -->0} can also be proven inductively.

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

P(n): d(nx)/dx=(ln n)n2 for all integer n.

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.

[cookiemonster]

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).

The binomial expansion is simply the taylor series of (x + 1)^n, is it not?

cookiemonster

[HallsofIvy]

The binomial expansion is simply the taylor series of (x + 1)^n, is it not?

cookiemonster

Therefore what? A power series expansion of any function is its Taylor series. My point was that it doesn't necessarily "involve the derivative". The fact that the coefficients are related to the derivative means that, if one has another way of finding a power series, one can use that to find the derivative- the whole point of this thread.

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+ x2+ ... by using the sum of a geometric series (known long before the calculus) and from that deduce that f '(0)= 1, f"(0)= 2, and, in general, f(n)(0)= n!.

[troutman]

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.

[Gokul43201]

\frac {d} {dx} 2^x (ln2) = (ln2) \frac {d} {dx} 2^x = (ln2)^2 2^x= \frac {d^2} {dx^2} 2^x

since ln2 is a constant, independent of x.

This can be repeated any number of times to give :

\frac {d^n} {dx^n} 2^x = (ln2)^n 2^x

[cair0]

\frac{dy}{dx} \left( \frac{1}{y} \right) = \ln (a)

\frac{dy}{dx} = y \ln(a)




you can do that?

[AKG]

cair0

Perhaps you misunderstood what Zurtex wrote:

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

[jatin9_99]

let y=2 ^ x
then log y = log 2^x
log y= x . log2

now, u can easily do this

[mathwonk]

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