Proof d/dx e^x=e^x using substitution

[Jkohn]

Homework Statement

Proof d/dx e^x=e^x, use e=limit (1+1/h)^h h->infinity

Show how that implies d/dx e^x=e^x

t

Homework Equations

The Attempt at a Solution

Ive tried using chain rule--wasnt accepted
Also, I did e=(1+h)^[1/h]-->e^h=1+h, then reduced to e^x, still didnt accept

How can I prove it using limit substitution? He told me to google finding limits by substitution then he said e^h-1=U

 

[Mentallic]

So have you tried taking the derivative of e^x in the limit form?

\frac{d}{dx}e^x=\frac{d}{dx}\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{hx}

 

[Jkohn]

So have you tried taking the derivative of e^x in the limit form?

\frac{d}{dx}e^x=\frac{d}{dx}\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{hx}

Ive attempted, but its a limit inside a limit?

 

[Mentallic]

Ive attempted, but its a limit inside a limit?

Aren't you allowed to use the derivative rules? \frac{d}{dx}a^{cx}=c\ln(a)a^{cx}

 

[Jkohn]

Aren't you allowed to use the derivative rules? \frac{d}{dx}a^{cx}=c\ln(a)a^{cx}

Exactly what i did when he first said, its obvious, but he wants mathematical reasoning using limits..

 

[klondike]

(e^x)'=\lim_{\delta\to 0}\dfrac{e^{x+\delta}-e^{x}}{\delta}=e^{x}\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}
Now you want to show
\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}=1
And you are given that
\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{h}=e
That's very close. Don't forget the hint you are given: letting u=e^h-1.

 

[Mentallic]

Exactly what i did when he first said, its obvious, but he wants mathematical reasoning using limits..

Hmm... ok...

Can you clarify what he meant by e^h-1=U?

What about if we tried to just go back to the basics. So we'll use any positive number a,

\frac{d}{dx}a^x=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}

=\lim_{h\to 0}\frac{a^xa^h-a^x}{h}

=a^x\lim_{h\to 0}\frac{a^h-1}{h}

Now, we're interested in the case where the limit is equal to 1,

\lim_{h\to 0}\frac{a^h-1}{h}=1

\lim_{h\to 0}a^h-1=\lim_{h\to 0} h

\lim_{h\to 0}a^h =\lim_{h\to 0}1+h

Can you see where this is heading?

edit: klondike beat me to it.

 

[klondike]

Now, we're interested in the case where the limit is equal to 1,
\lim_{h\to 0}\frac{a^h-1}{h}=1

uh-oh, this doesn't sound right.

 

[Jkohn]

Hmm... ok...

Can you clarify what he meant by e^h-1=U?

What about if we tried to just go back to the basics. So we'll use any positive number a,

\frac{d}{dx}a^x=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}

=\lim_{h\to 0}\frac{a^xa^h-a^x}{h}

=a^x\lim_{h\to 0}\frac{a^h-1}{h}

Now, we're interested in the case where the limit is equal to 1,

\lim_{h\to 0}\frac{a^h-1}{h}=1

\lim_{h\to 0}a^h-1=\lim_{h\to 0} h

\lim_{h\to 0}a^h =\lim_{h\to 0}1+h

Can you see where this is heading?

edit: klondike beat me to it.

Put that in my "attempts" he says that e^h ≠ (1+h) because its a ≈ and not =, I even told him under a certain domain "its true" he won't accept lol..

 

[Jkohn]

(e^x)'=\lim_{\delta\to 0}\dfrac{e^{x+\delta}-e^{x}}{\delta}=e^{x}\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}
Now you want to show
\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}=1
And you are given that
\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{h}=e
That's very close. Don't forget the hint you are given: letting u=e^h-1.

So what exactly should I be substituting??

 

[Mentallic]

uh-oh, this doesn't sound right.

Actually, yeah, you're right. We should be starting off with the limit expression for e.

 

[klondike]

let \frac{1}{h}=e^{\delta}-1 \quad;; h\to \infty \quad as \quad \delta \to 0 then write δ in terms of h, and plug them back in, and you are all set.

So what exactly should I be substituting??

 

[Jkohn]

(e^x)'=\lim_{\delta\to 0}\dfrac{e^{x+\delta}-e^{x}}{\delta}=e^{x}\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}
Now you want to show
\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}=1
And you are given that
\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{h}=e
That's very close. Don't forget the hint you are given: letting u=e^h-1.

let \frac{1}{h}=e^{\delta}-1 \quad;; h\to \infty \quad as \quad \delta \to 0 then write δ in terms of h, and plug them back in, and you are all set.

Im a bit confused here. Where are you getting the (1/h)=e^δ−1 from?

 

[klondike]

Because e^δ−1 goes zero as δ goes zero. You want some quantity goes zero as something goes zero or, more conveniently as something goes infinity as you are given a known limit as its independent variable goes infinity.

Hence, it's mostly conveniently to set the quantity to 1/h as h approach infinity. It satisfies both condition and closely resemble what you are given.
You can, of course first set y=e^δ−1 as well.

Im a bit confused here. Where are you getting the (1/h)=e^δ−1 from?

 

[Jkohn]

how do I put δ in terms of h..where am I plugging it in??
EDIT: SORRY I realized h=1/(e^δ - 1)

do I insert that h into the e= limit (1+ 1/h)^h ??

thanks

 

[klondike]

e^{\delta}-1=\frac{1}{h}
\delta=ln(1+\frac{1}{h})
And put these 2 into it. And recall that aln(x)=ln(x^a). I'm sure you will figure it out.

 

[Jkohn]

mhm I am doing this:

lim e^δ−1/ δ =1

e^δ - 1= 1/h
δ=ln(1+ 1/h)

so I get: limit [(1/h)]/ [ln(1+(1/h))] h-->infinity

Im getting 1

 

[klondike]

Now go back to check post #6. Isn't that what you're trying to prove?

mhm I am doing this:

lim e^δ−1/ δ =1

e^δ - 1= 1/h
δ=ln(1+ 1/h)

so I get: limit [(1/h)]/ [ln(1+(1/h))] h-->infinity

Im getting 1

 

[Jkohn]

ohhhhhh so 1=e^x holy ****!

 

[Jkohn]

well 1 implies it..wow so cool