# Homework Help: Hyperbolic differentiation

1. Oct 5, 2011

### nayfie

Hello again :)

I get the feeling I'm missing some kind of 'trick', as this is proving a very difficult question :(

I'll write out my frustration below;

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

Find f'(x) if f(x) = $\int^{cosh(x^{2})}_{0} tanh(t^2)dt$

2. Relevant equations

---

3. The attempt at a solution

My idea was to find the value of the integral (using integration by parts), simplify it and then differentiate it.

However, this method seems rather time consuming (and difficult).

Am I missing something?

2. Oct 5, 2011

### flyingpig

Fundamental Theorem of Calculus. What does it say? I am not talking about evaluating a definite integral

3. Oct 5, 2011

### nayfie

You mean that $\frac{d}{dx} \int^{x}_{a} f(t)dt = f(x)$?

Does this imply that $\frac{d}{dx} \int^{cosh(x^{2})}_{0} tanh(t^2)dt = tanh(cosh^2(x^2))$?

4. Oct 5, 2011

### flyingpig

What's the derivative of cosh(x^2)?

5. Oct 5, 2011

### vela

Staff Emeritus
Not quite. The statement of the fundamental theorem has an x as the upper limit. You have a function of x as the upper limit. That makes a difference.

6. Oct 5, 2011

### Staff: Mentor

Yes, that's what flyingpig is talking about.
No - you're neglecting to use the chain rule.

$$\frac{d}{dx} \int^{u(x)}_{a} f(t)dt = \frac{d}{du}\int^{u(x)}_{a} f(t)dt * \frac{du}{dx} = f(u) * \frac{du}{dx}$$

7. Oct 5, 2011

### SammyS

Staff Emeritus

Suppose the anti-derivative of tanh(t2) is G(t), i.e. G'(t) = tanh(t2).

For you problem this means that f(x) = G(cosh(x2)) - G(0).

What does that give you for f '(x) ? (Don;t forget to use the chain rule.)

8. Oct 6, 2011

### nayfie

First of all, thank you all for the contributions :)

I think I have arrived at an answer! Let me know if I've made a mistake.

-----------------------------------------------------------------

$f(x) = \int^{cosh(x^{2})}_{0} tanh(t^{2})dt$, let $tanh(t^2) = g(t)$

$f(x) = G(cosh(x^{2})) - G(0)$

$f(x) = G(cosh(x^{2}))$

$f'(x) = G'(cosh(x^{2})) = g(cosh(x^{2}))$

$f'(x) = tanh(cosh^{2}(x^{2})).sinh(x^{2})$

9. Oct 6, 2011

### I like Serena

You're on the right path!

But I'm afraid you haven't applied the chain rule properly yet.
This means that f'(x) = G'(t(x)) t'(x).
So you 4th line doesn't follow from the 3rd.

10. Oct 6, 2011

### nayfie

Actually now that I look at it again I think the chain rule needs to be applied twice?

$f(x) = G(cosh(x^{2}))$

$f'(x) = G'(cosh(x^{2})) = g(cosh(x^{2})).sinh(x^2).2x$

$f'(x) = tanh(cosh^{2}(x^{2})).sinh(x^{2}).2x$

It's been a long day :(

11. Oct 6, 2011

### I like Serena

Well, your day just came into fruition!
That's it!

Last edited: Oct 6, 2011
12. Oct 6, 2011

### nayfie

:)

Thank you all very much for the help!

13. Oct 6, 2011

### vela

Staff Emeritus
You shouldn't assume G(0) = 0. It turns out when you differentiate, it goes away, so it doesn't matter in the end. But you shouldn't simply erase it.