Can someone tell me how to figure out the answer to this question?

Question

the derivative of

tanh(x/y) = arctanh(y/x)

There IS NOTHING WRONG with the question, this is correct, im not on drugs or dislexic. Does someone know how to go about getting the derivative.

Answer 1

It helps to know some standard integrals and derivatives.
1) d/dx [atanh(x/a)] = a / (a^2 - x^2)
2) d/dx [tanh (x/a)] = 1/a * sech^2 (x/a)

With that let's differentiate implicitely
d/dx (LHS) = 1/y * sech^2 (x/y) * [y - x*dy/dx] / y^2
= [y - x*dy/dx] *sech^2 (x/y) / y^3

d/dx (RHS) = 1 / (1 - y^2/x^2) * [x*dy/dx - y] / x^2
= [x*dy/dx - y] * 1 / (x^2 - y^2)

Notice the common term on the LHS and RHS.
Rearranging,
(y - x*dy/dx) * [sech^2(x/y) / y^3 + x^2 / (x^2 - y^2) ] = 0

The only way for the LHS = RHS is for
y - x*dy/dx = 0
dy/dx = y/x

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This may be the equation of a hyperbola-like function.
By def'n -1 < tanh(x/y) < 1
So we must restrict the RHS to that range.
This means that -0.7616 < y/x < +0.7616
where 0.7616 = tanh(1)
So -tanh(1) * x < y < +tanh(1) * x
Those two lines y = +/- tanh(1) * x form the asymptotes of the "hyperbola". As x and y tend to infinity, the derivative dy/dx tends to +/- tanh(1).

Answer 2

Let´s take the derivative of both sides
1/ch^2(x/y)* 1/y^2 *(y-xy´)=1/2[ln (1+y/x)/(1-y/x)]´=
1/2[ln(x+y)/(x-y)]´= 1/2(x-y)/(x+y)*1/(x-y)^2 *[ (x-y)(1-y')-(x+y)(1-y´)]
From the first and the last you can calculate y´

Answer 3

you need to use the product rule.

ex: d/dx(x*y) = y*dy/dx

once you get the derivative of both sides, just solve for dy/dx (this is assuming that you are looking for dy/dx)