# 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?