1. ## tricky derivative?

Find the derivative of

2. We use the rule that

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

for $\displaystyle a>0$.

3. Originally Posted by Scott H
We use the rule that

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

for $\displaystyle a>0$.
yea I've been doing that but I'm not getting the right answer.. Must be doing something wrong.

4. Originally Posted by tbenne3
Find the derivative of
$\displaystyle \ln{y}=4^x\ln{x}$

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

Are you able to finish from there?

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

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

and differentiate using the Chain Rule.

6. Originally Posted by ione
$\displaystyle \ln{y}=4^x\ln{x}$

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

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

7. I recommend Scott's method.

Change the function to:

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

Then you differentiate:

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

All you use now is the product rule.

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

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

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

8. Originally Posted by Aryth
I recommend Scott's method.

Change the function to:

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

Then you differentiate:

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

All you use now is the product rule.

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

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

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