[Womble548]

This one is bothering me, I get a different answer to the back of the book

y = artanh (x^2)

This is what I have done:

tanh y = x^2
sech^2 y dy/dx = 2x
(1 - tanh^2 y) dy/dx = 2x
dy/dx = 2x/(1 - x^4)

The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

Can anyone help me here?

[RichE]

Originally Posted by Womble548

This one is bothering me, I get a different answer to the back of the book

y = artanh (x^2)

This is what I have done:

tanh y = x^2
sech^2 y dy/dx = 2x
(1 - tanh^2 y) dy/dx = 2x
dy/dx = 2x/(1 - x^4)

The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

Can anyone help me here?

I think the book's wrong

[Aitch]

Originally Posted by Womble548

This one is bothering me, I get a different answer to the back of the book

y = artanh (x^2)

This is what I have done:

tanh y = x^2
sech^2 y dy/dx = 2x
(1 - tanh^2 y) dy/dx = 2x
dy/dx = 2x/(1 - x^4)

The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

Can anyone help me here?

You're right. I've marked this one question [34] from Ex. 2A with a cross. I think the rest of the exercise is OK.

Aitch

[Aitch]

Originally Posted by Aitch

You're right. I've marked this one question [34] from Ex. 2A with a cross. I think the rest of the exercise is OK.

Aitch

..apart from the answer to question 60 of course...

[You may have a later edition than mine with the error corrected.]

Aitch

[Womble548]

Thanks, thought so

[animal555]

Originally Posted by Womble548

This one is bothering me, I get a different answer to the back of the book

y = artanh (x^2)

This is what I have done:

tanh y = x^2
sech^2 y dy/dx = 2x
(1 - tanh^2 y) dy/dx = 2x
dy/dx = 2x/(1 - x^4)

The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

Can anyone help me here?

Just thought Id add, the way you have done this question is okay but there is a quicker way if you have the differentiation of arctanh x given in yre formula booklet:

d(arctanh x)/dx = 1/(1 - x^2)

Therefore, to find y = arctanh (x^2) you simply substitute x^2 for x in the line above and mutiply this by the d(x^2)/dx as we must apply the chain rule:

d(arctanh x^2)/dx = 1/(1-(x^2)^2) x d(x^2)/dx

= 1/(1-x^4) x 2x
= 2x/(1-x^4)

I think this is the way yre textbook is trying to do the question but they forgot to square the x^2 when substituting it for x in d(arctanh x)/dx = 1/(1 - x^2)