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Differentiating exponential functions

Petrus

Well-known member
Feb 21, 2013
739
How do i derivate e^e^x (I don't know how to type it on latex but here you can se what i mean e^e^x - Wolfram|Alpha Results basicly dont know how I shall think
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: First and second derivate

If you have something of the form \(\displaystyle e^{f(x)}\) what is the derivative ?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

If you have something of the form \(\displaystyle e^{f(x)}\) what is the derivative ?
\(\displaystyle e^{f(x)}*f'(x)\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

How do i derivate e^e^x (I don't know how to type it on latex but here you can se what i mean e^e^x - Wolfram|Alpha Results basicly dont know how I shall think
Hi Petrus! :)

The chain rule says that the derivative of $f(g(x))$ is $f'(g(x))g'(x)$.
Now pick $f(y)=e^y$ and $g(x)=e^x$.
Can you apply the chain rule to $e^{e^x}$ (written in latex as e^{e^x}).
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

Petrus

Well-known member
Feb 21, 2013
739

chisigma

Well-known member
Feb 13, 2012
1,704
Re: First and second derivate

Some of most 'horrors' I have seen in my life are the so called 'towers of powers'... when I see an expression like $\displaystyle e^{e^{x}}$ I don't undestand if it is $\displaystyle e^{(e^{x})}$ or $\displaystyle (e^{e})^{x}$ (Headbang)...

Kind regards

$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

Some of most 'horrors' I have seen in my life are the so called 'towers of powers'... when I see an expression like $\displaystyle e^{e^{x}}$ I don't undestand if it is $\displaystyle e^{(e^{x})}$ or $\displaystyle (e^{e})^{x}$ (Headbang)...

Kind regards

$\chi$ $\sigma$
Hello!
I think this is hilirious because I got one problem like thatimage.jpg
anyone can give me advice on that aswell. I am suposed to derivate it once.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

Hello!
I think this is hilirious because I got one problem like thatView attachment 681
anyone can give me advice on that aswell. I am suposed to derivate it once.
So... how far do you get?
Do you know what the derivative of $2^y$ is?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

So... how far do you get?
Do you know what the derivative of $2^y$ is?
If u mean 2^x and derivate is ln(2)•2^x
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884

Petrus

Well-known member
Feb 21, 2013
739

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

Im really confused on this because its alot Power up to. Can you give me a advice
Okay.
Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.
Can you apply the chain rule?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

Okay.
Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.
Can you apply the chain rule?
I need first help how I shall think when i derivate $3^{x^2}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

I need first help how I shall think when i derivate $3^{x^2}$
Pick $f(y)=3^y$ and $g(x)=x^2$ this time round.
Can you apply the chain rule to that?
 

Petrus

Well-known member
Feb 21, 2013
739

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

$ln(3)3^y2x$
Almost.
When substituting in $f'(g(x))\cdot g'(x)$, you're supposed to replace all occurrence of $y$ in $f'(y)$ by $g(x)$.
That is, you're supposed to replace $y$ in $\ln(3) 3^y$ by $x^2$.
Can you do that?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

Almost.
When substituting in $f'(g(x))\cdot g'(x)$, you're supposed to replace all occurrence of $y$ in $f'(y)$ by $g(x)$.
That is, you're supposed to replace $y$ in $\ln(3) 3^y$ by $x^2$.
Can you do that?
$ln(3)3^{x^2}2x$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

$ln(3)3^{x^2}2x$
Good!
Let's go back to the original problem.
Can you apply the chain rule now?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

Okay.
Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.
Can you apply the chain rule?
(im replying to this cause i wanna use it as a mall)
I hope this trick work, I subsitate $x^2$ as c and $c'=2x$
$ln(2)2^yln(3)3^c*c'$
$ln(2)2^yln(3)3^{x^2}2x$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

(im replying to this cause i wanna use it as a mall)
I hope this trick work, I subsitate $x^2$ as c and $c'=2x$
$ln(2)2^yln(3)3^c*c'$
$ln(2)2^yln(3)3^{x^2}2x$
Looks good... almost there...
There is an $y$ left that should still be replaced by the $g(x)$...
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

Looks good... almost there...
There is an $y$ left that should still be replaced by the $g(x)$...
Hello,
Could you give me latex code for like 2^x^(y^2)
I cant replace that y with $3^{x^2}$ with latex I get error
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: First and second derivate

Hello,
Could you give me latex code for like 2^x^(y^2)
I cant replace that y with $3^{x^2}$ with latex I get error
You need curly braces {} to group symbols in latex.

So 2^x^(y^2) is 2^{x^{y^2}} which looks like $2^{x^{y^2}}$.
By adding a couple of \displaystyle directives we get $2^{\displaystyle x^{\displaystyle y^2}}$, which is more readable.

We need to be careful with these towers, because they look like the are on the verge of falling over. ;)
For safety we could keep them on the ground, like 2^(x^(y^2)).
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Re: First and second derivate

Nice tip I like Serena with the double use of \displaystyle! :)

You can even play around with the \hspace{} command and get it a little more readable. Here's what 2^{\hspace{.4 mm}\displaystyle x^{\hspace{.4 mm} \displaystyle y^2}} renders.

$2^{\hspace{.4 mm}\displaystyle x^{\hspace{.4 mm} \displaystyle y^2}}$ versus $2^{\displaystyle x^{\displaystyle y^2}}$
 

Petrus

Well-known member
Feb 21, 2013
739
Re: First and second derivate

You need curly braces {} to group symbols in latex.

So 2^x^(y^2) is 2^{x^{y^2}} which looks like $2^{x^{y^2}}$.
By adding a couple of \displaystyle directives we get $2^{\displaystyle x^{\displaystyle y^2}}$, which is more readable.

We need to be careful with these towers, because they look like the are on the verge of falling over. ;)
For safety we could keep them on the ground, like 2^(x^(y^2)).
$ln(2)2^{\displaystyle 3^{\displaystyle x^{\displaystyle2}}}ln(3)3^{\displaystyle x^2}2x$
Thanks for helping me and guiding me! Thanks for taking your time!:) I just learned some latex and improved my knowledge about Chain rule! cheer!(d)(Yes)(Muscle) :cool:
I owe you for the help! just (Phone)and I will invite you for some (Pizza)