Differentiate Exponential Functions

1. Sep 2, 2007

rocomath

this is an even problem of my homework so there is no answer

i think i did it right, i would just like a thumbs up/down ... thanks

$$y=2^{3^{x^{2}}}}$$

$$\ln{y}=\ln2^{3^{x^{2}}}}$$

$$\ln{y}=3^{x^{2}}}\times\ln2$$

$$\frac{y'}{2^{3^{x^{2}}}}}=\ln2\times3^{x^{2}}}\times\ln3\times2x$$

$$y'=2x\times2^{3^{x^{2}}}}\times3^{x^{2}}}\ln2\times\ln3$$

Last edited: Sep 2, 2007
2. Sep 2, 2007

rock.freak667

thumbs up..it is correct

3. Sep 2, 2007

olgranpappy

...there is no answer? That's deep, man.

4. Sep 2, 2007

bob1182006

thumbs up

but a faster method is to do the chain rule and remember that:

$$y=a^x$$
$$y'=a^xlna$$

5. Sep 2, 2007

olgranpappy

hmm... that's not actually too useful here, is it?

Perhaps more useful is something along the lines of:

If
$$y(x)=a^{f(x)}$$

then
$$y'(x)=a^{f(x)}\ln(a)f'(x)$$

6. Sep 2, 2007

i like that

7. Sep 2, 2007

thanks

8. Sep 3, 2007

Hurkyl

Staff Emeritus
Of course, you are doing exactly what bob said to do.

9. Sep 3, 2007

rocomath

so i was looking over my hw and found a few mistakes on some of them (even problems)

i'm suppose to simplify this problem.

1st one

$$e^{x+\ln{x}}$$

$$e^{x}\times e^{\ln{x}}$$

$$xe^{x}$$ should this be my final answer?

$$e^{x^{2}}$$ can i do this step or can i not assume it is was a power of?

Last edited: Sep 3, 2007
10. Sep 3, 2007

olgranpappy

No. His suggestion was a special case of mine having f(x)=x.

...
......

11. Sep 3, 2007

olgranpappy

Stop there! The next line does not follow

Nooooo. that is not a property of the exponential. You are confusing the properties of exponentials and logs...

12. Sep 3, 2007

Hurkyl

Staff Emeritus
He said to use the chain rule. What you wrote is what you get when you combine the chain rule with (a^x)' = a^x ln a

13. Sep 3, 2007

olgranpappy

So what? You said "exactly", but it was not "exactly what bob said to do".

I'm just pointing that out. Don't you agree?