Question

If y=cosh(4tanh¯¹x) then dy/dx=​

Answer 1

If you're like me and don't remember hyperbolic identities (especially involving inverse functions) off the top of your head, recall the definitions of the hyperbolic cosine and sine:

$\cosh(x) = \dfrac{e^x + e^{-x}}2 \\\\ \sinh(x) = \dfrac{e^x - e^{-x}}2$

Then differentiating yields

$\dfrac{\mathrm d(\cosh(x))}{\mathrm dx} = \dfrac{e^x-e^{-x}}2 = \sinh(x) \\\\ \dfrac{\mathrm d(\sinh(x))}{\mathrm dx} = \dfrac{e^x+e^{-x}}2 = \cosh(x)$

so that by the chain rule, if

$y = \cosh\left(4\tanh^{-1}(x)\right)$

then

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d\left(\cosh\left(4\tanh^{-1}(x)\right)\right)}{\mathrm dx} = \sinh\left(4\tanh^{-1}(x)\right) \dfrac{\mathrm d\left(4\tanh^{-1}(x)\right)}{\mathrm dx}$

Now, let $z=4\tanh^{-1}(x)$, so that (•) $\tanh\left(\frac z4\right) = x$.

Recall that

$\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)} = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

and so the derivative of tanh(x) is

$\dfrac{\mathrm d(\tanh(x))}{\mathrm dx} = \dfrac{(e^x+e^{-x})(e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x})^2} \\\\ = \dfrac{4\cosh^2(x)-4\sinh^2(x)}{4\cosh^2(x)} \\\\ = 1-\tanh^2(x) = \mathrm{sech}^2(x)$

where the last equality follows from the hyperbolic Pythagorean identity,

$\cosh^2(x)-\sinh^2(x) = 1 \implies 1-\tanh^2(x) = \mathrm{sech}^2(x)$

Differentiating both sides of (•) implicitly with respect to x gives

$\dfrac{\mathrm d\left(\tanh\left(\frac z4\right)\right)}{\mathrm dx} = \dfrac{\mathrm d(x)}{\mathrm dx} \\\\ \mathrm{sech}^2\left(\dfrac z4\right)\dfrac{\mathrm d\left(\frac z4\right)}{\mathrm dx} = 1 \\\\ \dfrac14 \mathrm{sech}^2\left(\dfrac z4\right) \dfrac{\mathrm dz}{\mathrm dx} = 1 \\\\ \dfrac{\mathrm dz}{\mathrm dx} = 4\cosh^2\left(\dfrac z4\right) \\\\ \dfrac{\mathrm dz}{\mathrm dx} = 4\cosh^2\left(\tanh^{-1}(x)\right)$

So, the derivative we want is the somewhat messy expression

$\boxed{\dfrac{\mathrm dy}{\mathrm dx} = 4\sinh\left(4\tanh^{-1}(x)\right) \cosh^2\left(\tanh^{-1}(x)\right)}$

and while this could be simplified into a rational expression of x, I would argue for leaving the solution in this form considering how y is given in this form from the start.

In case you are interested, we have

$\cosh\left(\tanh^{-1}(x)\right) = \dfrac1{\sqrt{1-x^2}}$

and you can instead work on differentiating that; you would end up with

$\dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{16x^3+16x}{(x^2-1)^3}$