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

[uzman1243]

Homework Statement

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

Homework Equations

see above

The Attempt at a Solution

Im trying to use the chain rule here but I can't even get the first step. Can you guide me?

 

[Dick]

Homework Statement

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

Homework Equations

see above

The Attempt at a Solution

Im trying to use the chain rule here but I can't even get the first step. Can you guide me?

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.

 

[uzman1243]

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.

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

But how do I use the chain rule here?

 

[Dick]

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

But how do I use the chain rule here?

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?

 

[HallsofIvy]

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.

 

[uzman1243]

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?

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

 

[Dick]

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

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.

 

[uzman1243]

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.

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

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

 

[Dick]

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

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

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?

 

[uzman1243]

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?

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

 

[Dick]

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

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