# Derivative and Integral Question

1. May 13, 2008

### Oneiromancy

1. y = 3^$$^{}log_{}_{}_2^{}(t)$$ Find dy/dx ... I tried using logarithmic differentiation but that didn't work.

2. $$\int x^{2x}(1 + ln x)dx\$$ ... I set u = $$x^{2x}$$ but my du didn't quite work out right.

2. May 13, 2008

### Dick

a^b=e^(b*ln(a)). log_a(b)=ln(b)/ln(a). I don't think you are trying hard enough. Try again, and show us what you tried this time if you are still having problems.

3. May 13, 2008

### Tom Mattson

Staff Emeritus

Are you sure? Unless t is some function of x, that derivative is identically zero.

4. May 13, 2008

### Oneiromancy

I meant dy/dt. :)

I will try to solve these again a little later, thanks.

5. May 13, 2008

### lurflurf

use the power rule
(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'
or
[log(u^v)]'=v{[log(u)]'+log(u)[log(v)]'}

Last edited: May 13, 2008
6. May 13, 2008

### Dick

I wouldn't recommend you do that unless you are willing to carry those formulas around in your head for the rest of your life. You don't need them. Sorry, lurflurf. Did you really remember those, or did you derive them just now??

Last edited: May 13, 2008
7. May 14, 2008

### Oneiromancy

I understand how to get the answer now, thanks.

8. May 14, 2008

### lurflurf

I really remember those.
I think if you are going to remember 15-20 differentiation formulas that would be one of them.
If one recalls
(u^v)'=v*u^(v-1)*u' when v'=0 obvious since [x^a]'=a*x^(a-1)
(u^v)'=u^v*log(u)*v' when u'=0 obvious since [exp(a*x)]'=a*exp(a*x)
one knows by the chainn rule the sum generalizes to the case where neither is constant
it is actually easier to recall
(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'
that two seperate equations

One must decide how many calculus formulas to carry in ones head (or written on the back of ones hand) at different points in life.
That formula is more helful than say
[log(sin(a*x))]'=a*cot(a*x)
and less useful than
c'=0
in particular if I ever forget that formula it is easer to derive the general result when needed and apply it than to derive a special case.

9. May 14, 2008

### Dick

Interesting. I just do (u^v)'=[e^(ln(u)*v)]' and go from there. But you are quite right. The number and selection of calculus formulas you need to carry in your head is a personal choice. But I don't think I'd ever even seen those before.

10. May 15, 2008

### DavidWhitbeck

Just simplify your expression before taking the derivative.

Have you ever found the derivative of $$x^x$$ before? Knowing that could be quite helpful.