# Derivative of (tanh^-1(sinh(2x)))

1. Apr 11, 2014

### uzman1243

1. The problem statement, all variables and given/known data
Derivative of (tanh^-1(sinh(2x)))

2. Relevant equations
see above

3. The attempt at a solution
Im trying to use the chain rule here but I cant even get the first step. Can you guide me?

2. Apr 11, 2014

### Dick

tanh^(-1)(x) is also call arctanh(x). What's the derivative of that? If you don't know you could try and look it up.

3. Apr 11, 2014

### uzman1243

Yes I am aware of that. The derivative is:
1/(1-x^2)

But how do I use the chain rule here?

4. Apr 11, 2014

### Dick

The chain rule says the derivative of (f(g(x)))'=f'(g(x))*g'(x). f is arctanh and g(x) is cosh(2x). So?

5. Apr 11, 2014

### HallsofIvy

Staff Emeritus
Another way to do this is to change $y= tanh^{-1}(sinh(2x)$ to $tanh(y)= sinh(2x)$.

Differentiate both sides with respect to x (so the left side will be the derivative of tanh(y) with respect to y times dy/dx) and then solve for dy/dx.

6. Apr 12, 2014

### uzman1243

it would be:

(1/(1-x^2)) * sinh2x * 2cosh2x

Is that correct? I checked wolfram for the derivative and i got:
-4cosh(2x) / cosh(4x)-3

Im not sure how I could get my solution in that format?

Thank you so much for helping me out

Last edited: Apr 12, 2014
7. Apr 12, 2014

### Dick

If f is arctanh and g(x) is cosh(2x). Then just as f'(x)=1/(1-x^2), f'(g(x)) should be 1/(1-g(x)^2). NOT 1/(1-x^2). Don't worry too much about comparing with Wolfram. There are many different looking ways to write the same answer.

8. Apr 13, 2014

### uzman1243

so is this correct:
Derivative of (tanh^-1(sinh(2x))) is

(1/1-(2sinh(2x)^2) * 2cosh(2x) ?

9. Apr 13, 2014

### Dick

Ok, so the right hand factor should be g'(x)=(sinh(2x))'=2cosh(2x). That looks right. The rest should be f'(g(x))=arctanh'(g(x))=1/(1-g(x)^2)). That doesn't look right. What's wrong with it?

10. Apr 13, 2014

### uzman1243

oh a careless mistake.
is it:
(1/1-(sinh(2x)^2) * 2cosh(2x) ?

11. Apr 13, 2014

### Dick

Yes, it is. You should balance your parentheses more carefully, but I know what you mean.

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