# Need Calculus Help

1. Jan 13, 2005

### Yapper

I am studying for a Calculus test and I need some help withsome concepts...

1) The problem is find the derivatie of y= (integral a=1 b=sinx of) t^3 dt. I went into my book and found a formula for it. The derivative of (intergral a=constant b=x of) f(t) dt = f(x). So I got (sinx)^3 is this right? If so can someone explain this concept and provide a proof for it? My text book isn't good at explaining things...

2)The problem is to differentiate y=log base x (2x). So I chamged it to ln2x/lnx from there can i just use divison rule? u'v -uv'/v^2? in that case would it be (lnx/x-ln2x/x)/(lnx)^2?

2. Jan 13, 2005

### vincentchan

1: is your derivative repect to x or t? this make a huge different
2: you are perfectly correct....

3. Jan 13, 2005

### Yapper

y= (integral a=1 b=sinx of) t^3 dt, so the differential unit is t, since it is dt, im guessing, thats all the information i was given

4. Jan 13, 2005

### dextercioby

Then it's wrong.Compute the antiderivative of the integrand and apply the fundamental formula of Leibniz & Newton.

Daniel.

5. Jan 13, 2005

### Yapper

No clue what you just said...

6. Jan 13, 2005

### Yapper

How do I use LaTex to make it clearer?

7. Jan 13, 2005

### vincentchan

OK, your integrand is t^3 and the limit is from 1 to sinx.... right?
Here is how u do the derivative of this monster...
if it is repect to t.. do the integral and substitude 1 and sinx in.... and do the derivative
if it is repect to x.. do the integral and substitude 1 and sinx in.... and do the derivative

basically, what you gonna do is do the integral and substitude 1 and sinx in.... and do the derivative, OK?

but if the derivative is repect to t, the result is trivial......

8. Jan 13, 2005

### Yapper

I dont understand.... I do the integral and I get t^4/4 substitutre for sinx and do the derivative I get sinx^3cosx is that right?

9. Jan 13, 2005

### dextercioby

There's a compiler that will "translate" all code lines u type in the window in which u write the message.Use the function "preview post" to check formulas for typos.

The result is trivial.It can be shown that
$$\frac{d}{dx}{\int_{const.}^{f(x)} u(t) dt =u(f(x))f'(x)$$

I still reccomend to you to do the integration and then the differentiation and confront with the result my formula gives you.

Daniel.

10. Jan 13, 2005

### vincentchan

$$\frac{d}{dx}{\int_{const.}^{f(x)} u(t) dt =u(f(x)) \frac{d}{dx}f(x)$$

11. Jan 13, 2005

### Yapper

Oh ok... my books says g(x) = (intergral from a constant to x of) f(t) dt, then g'(x) = f(x)

12. Jan 13, 2005

### Yapper

Thanks for all the help!!! and is there a website or a help file that has the code?

Last edited: Jan 13, 2005
13. Jan 13, 2005

### cepheid

Staff Emeritus

14. Jan 13, 2005

### dextercioby

Yes,PF is one of them.Check out the "sticky" in the General Physics forum called "Intriducing LaTex typesetting".U'll find amog those posts a link to a ".pdf"file with the code.
Quicker version.Click (left) on one of the formulas in this (and any other one) thread and u'll be opened (if u don't have a pop-up blocker) a new window with the code for that specific formula and a link to the ".pdf" file with the code.

And it's "inputs"... :tongue2:

Daniel.