# Hyperbolic differentiation

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;

## Homework Statement

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

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## 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?

Related Calculus and Beyond Homework Help News on Phys.org
Fundamental Theorem of Calculus. What does it say? I am not talking about evaluating a definite integral

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

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

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

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

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

Mark44
Mentor
You mean that $\frac{d}{dx} \int^{x}_{a} f(t)dt = f(x)$?
Yes, that's what flyingpig is talking about.
Does this imply that $\frac{d}{dx} \int^{cosh(x^{2})}_{0} tanh(t^2)dt = tanh(cosh^2(x^2))$?
No - you're neglecting to use the chain rule.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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

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.)

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.

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

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

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

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

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

I like Serena
Homework Helper
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.

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 :(

I like Serena
Homework Helper
Well, your day just came into fruition!
That's it!

Last edited:
:)

Thank you all very much for the help!

vela
Staff Emeritus