- #1
uzman1243
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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?
uzman1243 said: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 said: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 said:Yes I am aware of that. The derivative is:
1/(1-x^2)
But how do I use the chain rule here?
Dick said: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?
uzman1243 said: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 said: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 said:so is this correct:
Derivative of (tanh^-1(sinh(2x))) is
(1/1-(2sinh(2x)^2) * 2cosh(2x) ?
oh a careless mistake.Dick said: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 said:oh a careless mistake.
is it:
(1/1-(sinh(2x)^2) * 2cosh(2x) ?
The derivative of tanh^-1(sinh(2x)) is 2sech^2(2x). This can be found by using the chain rule and the derivative of tanh^-1(x) = 1/(1-x^2).
The domain of tanh^-1(sinh(2x)) is all real numbers. This is because both tanh^-1(x) and sinh(2x) have a domain of all real numbers.
To find the critical points of tanh^-1(sinh(2x)), you can set the derivative equal to 0 and solve for x. This will give you the x-values of the critical points.
tanh^-1(sinh(2x)) is an odd function. This is because both tanh^-1(x) and sinh(2x) are odd functions, and the composition of two odd functions is also an odd function.
The limit of tanh^-1(sinh(2x)) as x approaches infinity is 1. This can be found by using the limit definition of tanh^-1(x) and the limit of sinh(2x) as x approaches infinity, which is also 1.