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#### Petrus

##### Well-known member

- Feb 21, 2013

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- Feb 21, 2013

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- Jan 17, 2013

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

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- Feb 21, 2013

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

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- #4

- Mar 5, 2012

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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}).

- Jan 17, 2013

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

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- Feb 21, 2013

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

- Feb 13, 2012

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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

$\chi$ $\sigma$

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- Feb 21, 2013

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Hello!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

$\chi$ $\sigma$

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.

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- #9

- Mar 5, 2012

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So... how far do you get?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.

Do you know what the derivative of $2^y$ is?

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- #10

- Feb 21, 2013

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If u mean 2^x and derivate is ln(2)•2^xSo... how far do you get?

Do you know what the derivative of $2^y$ is?

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- #11

- Mar 5, 2012

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

So...?

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- #12

- Feb 21, 2013

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Im really confused on this because its alot Power up to. Can you give me a adviceGood!

So...?

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- #13

- Mar 5, 2012

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

Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.

Can you apply the chain rule?

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- #14

- Feb 21, 2013

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I need first help how I shall think when i derivate $3^{x^2}$Okay.

Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.

Can you apply the chain rule?

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- #15

- Mar 5, 2012

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Pick $f(y)=3^y$ and $g(x)=x^2$ this time round.I need first help how I shall think when i derivate $3^{x^2}$

Can you apply the chain rule to that?

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- #16

- Feb 21, 2013

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$ln(3)3^y2x$Pick $f(y)=3^y$ and $g(x)=x^2$ this time round.

Can you apply the chain rule to that?

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- #17

- Mar 5, 2012

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Almost.$ln(3)3^y2x$

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?

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- #18

- Feb 21, 2013

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

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- #19

- Mar 5, 2012

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

Let's go back to the original problem.

Can you apply the chain rule now?

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- Feb 21, 2013

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(im replying to this cause i wanna use it as a mall)Okay.

Let's start by picking $f(y)=2^y$ and $g(x)=3^{x^2}$.

Can you apply the chain rule?

I hope this trick work, I subsitate $x^2$ as c and $c'=2x$

$ln(2)2^yln(3)3^c*c'$

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

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- #21

- Mar 5, 2012

- 8,884

Looks good... almost there...(im replying to this cause i wanna use it as a mall)

I hope this trick work, I subsitate $x^2$ as c and $c'=2x$

$ln(2)2^yln(3)3^c*c'$

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

There is an $y$ left that should still be replaced by the $g(x)$...

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- #22

- Feb 21, 2013

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Hello,Looks good... almost there...

There is an $y$ left that should still be replaced by the $g(x)$...

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

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- #23

- Mar 5, 2012

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

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)).

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- #24

- Jan 26, 2012

- 4,052

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 $2^{\displaystyle x^{\displaystyle y^2}}$

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- #25

- Feb 21, 2013

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$ln(2)2^{\displaystyle 3^{\displaystyle x^{\displaystyle2}}}ln(3)3^{\displaystyle x^2}2x$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)).

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