Could i get some help working out dy/dx of arccoth(x)

[inflames829]

Homework Statement

y=arccoth(x), find the derivative

Homework Equations

cothx - coshx/sinhx

The Attempt at a Solution

y = arccoth(x)
x = coth(y)
= cosh(y)/sinh(y)
quotient rule = (sinh(y)*sinh(y))-(cosh(y)*cosh(y))/(sinh(y))^2
= sinh^2(y) - cosh^2(y)/sinh^2(y)
x= -1/sinh^2(y)

this is what i got which is the derivative of coth, but no the inverse. Is there a way to go from here and rearrange it back to y =

 

[hunt_mat]

You're almost there! If y=argcoth(x) then x=costh(y), then from the chain rule:
$$1=\frac{dy}{dx}\frac{d}{dy}(\coth x)$$
Re-arrange to get dy/dx

 

[inflames829]

sorry but I'm still not getting it, can you give me another hint or show me please

 

[hunt_mat]

ok, you have:
$$y=\coth^{-1}x$$
Then as you have stated
$$x=\coth y$$
Differentiate this w.r.t. x and use the chain rule:
$$1=\frac{d}{dx}(\coth y)=\frac{d}{dy}(\coth y)\frac{dy}{dx}$$
You know how to compute the derivative of coth x, so...

 

[inflames829]

does that mean 1 = -1/sinh^2(y) * (coth(y))

 

[hunt_mat]

No it means that:
$$1=-\sinh^{2}y\frac{dy}{dx}$$

 

[Bohrok]

The Attempt at a Solution

y = arccoth(x)
x = coth(y)
= cosh(y)/sinh(y)
quotient rule = (sinh(y)*sinh(y))-(cosh(y)*cosh(y))/(sinh(y))^2
= sinh^2(y) - cosh^2(y)/sinh^2(y)
x= -1/sinh^2(y)

You differentiated the right side with respect to x, but not the left side. You should have
$$x = \frac{\cosh(y)}{\sinh(y)} \Rightarrow \frac{d}{dx}x = \frac{d}{dx}\frac{\cosh(y)}{\sinh(y)} \Rightarrow 1 = -\frac{1}{\sinh^2(y)}\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\sinh^2(y)$$

Since y = arccoth(x), we have
$$-\sinh^2y = -\sinh^2(arccoth(x)) = \frac{1}{-csch^2(arccoth(x))}$$

Then using the identity coth²x = 1 + csch²x ==> -csch²x = 1 - coth²x, we have

$$\frac{dy}{dx} = \frac{1}{-csch^2(arccoth(x))} = \frac{1}{1 -coth^2(arccoth(x))} = ~?$$

 

[inflames829]

thanks for that I have the answer now