# Homework Help: 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!