Differentiating exponential functions

[Petrus]

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 don't know how I shall think

 

[alyafey22]

Re: First and second derivate

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

 

[Petrus]

Re: First and second derivate

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

\(\displaystyle e^{f(x)}*f'(x)\)

 

[I like Serena]

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 don't 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}).

 

[alyafey22]

Re: First and second derivate

\(\displaystyle e^{f(x)}*f'(x)\)

Excellent , now put \(\displaystyle f(x)=e^x\) .

 

[Petrus]

Re: First and second derivate

Excellent , now put \(\displaystyle f(x)=e^x\) .

$e^{e^x}*e^x$

 

[chisigma]

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]

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 thatView attachment 681
anyone can give me advice on that aswell. I am suposed to derivate it once.

 

[I like Serena]

Re: First and second derivate

Hello!
I think this is hilirious because I got one problem like thathttps://www.physicsforums.com/attachments/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]

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

 

[I like Serena]

Re: First and second derivate

If u mean 2^x and derivate is ln(2)•2^x

Good!
So...?

 

[Petrus]

Re: First and second derivate

Good!
So...?

Im really confused on this because its a lot Power up to. Can you give me a advice

 

[I like Serena]

Re: First and second derivate

Im really confused on this because its a lot 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]

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 $3^{x^2}$

 

[I like Serena]

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]

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?

$ln(3)3^y2x$

 

[I like Serena]

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]

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?

$ln(3)3^{x^2}2x$

 

[I like Serena]

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]

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 want to use it as a mall)
I hope this trick work, I subsitate $x^2$ as c and $c'=2x$
$ln(2)2^yln(3)3^c*c'$
$ln(2)2^yln(3)3^{x^2}2x$

 

[I like Serena]

Re: First and second derivate

(im replying to this cause i want to 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]

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 can't replace that y with $3^{x^2}$ with latex I get error

 

[I like Serena]

Re: First and second derivate

Hello,
Could you give me latex code for like 2^x^(y^2)
I can't 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]

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]

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)
I owe you for the help! just (Phone)and I will invite you for some (Pizza)

 

[HallsofIvy]

By the way, you could have differentiated [tex]y= 2^{x^2}[/tex] by taking the logarithm of both sides:

[tex]\ln(y)= x^2\ln(2)[/tex]

Now, it is easy to differentiate both sides with respect to $x$:

[tex]\frac{1}{y}\frac{dy}{dx}= 2x\ln(2)[/tex]
[tex]\frac{dy}{dx}= 2xy\ln(2)= 2\ln(2)x 2^{x^2}[/tex]