# Homework Help: Fundamental Theorem of Calculus Pt 2, multivariable integration?

1. Jan 19, 2012

### mleeno

1. The problem statement, all variables and given/known data

problem in attachment

2. Relevant equations

3. The attempt at a solution

I can get f'(x) as sqrt(1 + (sinx)^2) and derive that to get the second derivative but as far as that I don't really get the concept behind this question

will y be another function I have to use chain rule with?
or is it just a variable and I plug in pi/6 normally?

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2. Jan 19, 2012

### SammyS

Staff Emeritus
What is $\displaystyle \frac{d}{dy}g(y)\,?$

3. Jan 19, 2012

### mleeno

g'(y) = f(y) ?
g'(y) = f(y)y'

4. Jan 19, 2012

### SammyS

Staff Emeritus
I intended to show the image of your problem in my previous post.

By the Fund. Thm. of Calc., Pt. 2: g'(y) = f(y). What you use for the independent variable is unimportant, so $g'(x)=f(x)\,.$

Then what is $g''(x)\,?$

BTW: Welcome to PF !

5. Jan 19, 2012

### HallsofIvy

$g(y)= \int_3^y f(x)dx$ is NOT a multiple variable function, it is a function of the single variable y.

The "fundamental theorem of Calculus" says that g'(y)= f(y). The second derivative of g is simply the first derivative of f.

$f(u)= \int_0^u\sqrt{1+ t^2}dt$ where $u= sin(x)$. Now apply both the fundamental theorem of Calculus and the chain rule.

6. Jan 19, 2012

### mleeno

AH so

simply f'(pi/6) = g''(pi/6)

thanks for the help everyone.

I got something like

f'(x) = (cosx)sqrt(1 + (sinx)^2)

plugging in f'(pi/6) = cos(pi/6)sqrt(1 + (sin(pi/6)^2))

which should be [sqrt(3)/2]*sqrt(5/4) ???