Fundamental Theorem of Calculus Pt 2, multivariable integration?

[mleeno]

Homework Statement

problem in attachment

Homework Equations

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?

 

[SammyS]

Homework Statement

problem in attachment

Homework Equations

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?

What is [itex]\displaystyle \frac{d}{dy}g(y)\,?[/itex]

 

[mleeno]

What is [itex]\displaystyle \frac{d}{dy}g(y)\,?[/itex]

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

I think I'm thinking about this too hard...

 

[SammyS]

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

I think I'm thinking about this too hard...

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 [itex]g'(x)=f(x)\,.[/itex]

Then what is [itex]g''(x)\,?[/itex]

BTW: Welcome to PF !

 

[HallsofIvy]

[itex]g(y)= \int_3^y f(x)dx[/itex] 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.

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

 

[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) ?