If Integral with Sine Limits What is Second Derivative?

1. Aug 27, 2014

Gwozdzilla

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

If f(x) = ∫sin x0 √(1+t2)dt and g(y) = ∫3y f(x)dx, find g''(pi/6)?

2. Relevant equations

FTC: F(x) = ∫f(x)dx
ab f(t)dt = F(b) - F(a)

Chain Rule:
f(x) = g(h(x))
f'(x) = g'(h(x))h'(x)

3. The attempt at a solution

I tried u-substition setting u = tan(x) for the first dirivative with the limits of sine, and it got really weird and bad. I also tried trigonometric substition and I got a similar bad and ugly answer.

g'(y) = f(x) = ∫0sin x √(1+t2)dt

How do I take the second derivative of g(y)?

2. Aug 27, 2014

haruspex

That doesn't really make any sense. In the definition of g, x is a 'dummy' variable. It doesn't exist outside of the integral. Have another go.

3. Aug 27, 2014

Zondrina

Apply the fundamental theorem to $g(y)$:

$$g'(y) = \frac{d}{dy} \int_3^y f(x) dx$$

4. Aug 27, 2014

Gwozdzilla

If g = ∫f(x), why isn't g' = f(x)? Is the second derivative of g not equal to the first derivative of f? If it doesn't, then I guess I have no clue where to start. Do I need to integrate √(1+t^2)dt?

5. Aug 27, 2014

Gwozdzilla

Maybe I don't understand how to apply FTC..

g'(y) = f(y) - f(3) ?

If that's true then I still don't know where to go from here...

6. Aug 27, 2014

Zondrina

You are forgetting the chain rule:

$$g'(y) = \frac{d}{dy} \int_3^y f(x) dx = f(y) \frac{d}{dy} [y] - f(3) \frac{d}{dy} [3] = f(y)$$

What is $f(y)$? Take the second derivative of $g(y)$ now.

7. Aug 27, 2014

Gwozdzilla

Why do I need to use the chain rule? Are y and 3 functions? How can you tell when you need to use the cahin rule?

Besides, if y is the variable, then isn't d/dy y just equal to 1 and d/dy 3 = 0? Does that make g'(y) = f(y)? Am I going in completely the wrong direction here?

8. Aug 27, 2014

Zondrina

The fundamental theorem stated in a general form is:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)$$

Quick proof:

$$\int_{a(x)}^{b(x)} f(t) dt = F(t) |_{a(x)}^{b(x)} = F(b(x)) - F(a(x))$$

Where $F(t)$ is an anti derivative of $f(t)$. Taking the derivative of the expression:

$$\frac{d}{dx} F(b(x)) - F(a(x)) = f(b(x)) b'(x) - f(a(x)) a'(x)$$

Where $f(t)$ is the derivative of $F(t)$ and we have applied the chain rule to the composition of the functions.

We applied this theorem just now to take the first derivative with respect to $y$ of $g(y)$. You find this is equal to $f(y)$. Write out what $f(y)$ is and then take the second derivative.

9. Aug 27, 2014

Gwozdzilla

Okay, I think I follow most of the way now... I understand how FTC works now.

I guess now I'm just confused about how and what to take the second derivative of:

Do I need to write ∫sin x0 √(1+t2)dt in terms of y or is there something more I can do with g'(y) by doing the chain rule again somehow?

10. Aug 27, 2014

Zondrina

Yes, you literally want to plug in $y$ for $x$ in $f(x)$. This will change the limits on the integral, namely you now have a $\sin(y)$. Then you need to take the derivative again to get $g''(y)$.

11. Aug 27, 2014

haruspex

Think about what differentiation means. g'(y) means you make a small change in y and observe the consequent change in g(y). Think of the integral as a sum. If you increase the upper limit slightly on ∫yf(x), making it ∫y+dyf(x), how much does the 'sum' increase by?

12. Aug 27, 2014

Gwozdzilla

Okay, let me try!

g'(y) = f(y) = ∫sin(y)0 sqrt(1+t2)dt

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

Do I do this one with FTC as well or should I do a u-subs with x = tan(u)?

If I try it with FTC...

f'(y) = √(1+sin2y) d/dy sin(y) - √(1 + 02) d/dy (0)

f'(y) = √(1+sin2y) cos(y) = g''(y)

g''(pi/6) = √(5/4)((√3)/2)

Is that right?

13. Aug 27, 2014

Zondrina

Everything looks good, it seems you can apply the theorem now.

You should spend some time asking yourself what the implications of this theorem are. Try to get a geometric grasp to aid in visualizing.

14. Aug 27, 2014

Gwozdzilla

Thank you very much for all of your help! I will see what I can do to figure out the implications and to better visualize it.

15. Aug 27, 2014

Ray Vickson

You wrote something that makes no sense: you should have $g'(y) = f(y)$, not $f(x)$. You cannot have unrelated $x$ and $y$ on opposite sides of the same equation!