# Differentiating exponential functions

#### Petrus

##### Well-known member
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
Re: First and second derivate

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

#### Petrus

##### Well-known member
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
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
Re: First and second derivate

$$\displaystyle e^{f(x)}*f'(x)$$
Excellent , now put $$\displaystyle f(x)=e^x$$ .

#### Petrus

##### Well-known member
Re: First and second derivate

Excellent , now put $$\displaystyle f(x)=e^x$$ .
$e^{e^x}*e^x$

#### chisigma

##### Well-known member
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 $\displaystyle (e^{e})^{x}$ ...

Kind regards

$\chi$ $\sigma$

#### Petrus

##### Well-known member
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 $\displaystyle (e^{e})^{x}$ ...

Kind regards

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
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
Re: First and second derivate

If u mean 2^x and derivate is ln(2)•2^x
Good!
So...?

#### Petrus

##### Well-known member
Re: First and second derivate

Good!
So...?
Im really confused on this because its alot Power up to. Can you give me a advice

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
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
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
Re: First and second derivate

Pick $f(y)=3^y$ and $g(x)=x^2$ this time round.
Can you apply the chain rule to that?
$ln(3)3^y2x$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
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
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
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
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
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
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

Staff member
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
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)
I owe you for the help! just and I will invite you for some