tricky derivative?

**tbenne3** · February 24th 2010, 01:46 PM

Find the derivative of

**Scott H** · February 24th 2010, 02:09 PM

We use the rule that

$a^r=e^{r\ln a}$

for $a>0$ .

**tbenne3** · February 24th 2010, 02:31 PM

Originally Posted by Scott H

We use the rule that

$a^r=e^{r\ln a}$

for $a>0$ .

yea I've been doing that but I'm not getting the right answer.. Must be doing something wrong.

**ione** · February 24th 2010, 03:55 PM

Originally Posted by tbenne3

Find the derivative of

$\ln{y}=4^x\ln{x}$

$\frac{1}{y}\,\frac{dy}{dx}=4^x\,\frac{1}{x}+\ln{x} \,(4^x\,\ln{4})$

Are you able to finish from there?

**Scott H** · February 24th 2010, 04:44 PM

I imagined that that differential equation would be a bit tricky. We can rewrite the function as

$y=x^{4^x}=e^{4^x\ln x}$

and differentiate using the Chain Rule.

**tbenne3** · February 25th 2010, 07:10 AM

Originally Posted by ione

$\ln{y}=4^x\ln{x}$

$\frac{1}{y}\,\frac{dy}{dx}=4^x\,\frac{1}{x}+\ln{x} \,(4^x\,\ln{4})$

Are you able to finish from there?

honestly.. no.. our teacher doesn't really know how to teach and I haven't had time to sit down and teach it to myself yet

**Aryth** · February 25th 2010, 07:45 AM

I recommend Scott's method.

Change the function to:

$y = e^{4^x\ln{x}}$

Then you differentiate:

$y' = e^{4^x\ln{x}}\frac{d}{dx}[4^x\ln{x}]$

All you use now is the product rule.

$y' = e^{4^x\ln{x}}\left(4^x\log{4}\ln{x} + \frac{4^x}{x}\right)$

We remember that: $e^{4^x\ln{x}} = x^{4^x}$

$y' = 4^xx^{4^x}\left(\ln{x}\log{4} + \frac{1}{x}\right)$

**tbenne3** · February 25th 2010, 11:46 AM

Originally Posted by Aryth

I recommend Scott's method.

Change the function to:

$y = e^{4^x\ln{x}}$

Then you differentiate:

$y' = e^{4^x\ln{x}}\frac{d}{dx}[4^x\ln{x}]$

All you use now is the product rule.

$y' = e^{4^x\ln{x}}\left(4^x\log{4}\ln{x} + \frac{4^x}{x}\right)$

We remember that: $e^{4^x\ln{x}} = x^{4^x}$

$y' = 4^xx^{4^x}\left(\ln{x}\log{4} + \frac{1}{x}\right)$

thanks

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