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:
[tex]\cosh(x) = \dfrac{e^x + e^{-x}}2 \\\\ \sinh(x) = \dfrac{e^x - e^{-x}}2[/tex]
Then differentiating yields
[tex]\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)[/tex]
so that by the chain rule, if
[tex]y = \cosh\left(4\tanh^{-1}(x)\right)[/tex]
then
[tex]\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}[/tex]
Now, let [tex]z=4\tanh^{-1}(x)[/tex], so that (•) [tex]\tanh\left(\frac z4\right) = x[/tex].
Recall that
[tex]\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)} = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}[/tex]
and so the derivative of tanh(x) is
[tex]\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)[/tex]
where the last equality follows from the hyperbolic Pythagorean identity,
[tex]\cosh^2(x)-\sinh^2(x) = 1 \implies 1-\tanh^2(x) = \mathrm{sech}^2(x)[/tex]
Differentiating both sides of (•) implicitly with respect to x gives
[tex]\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)[/tex]
So, the derivative we want is the somewhat messy expression
[tex]\boxed{\dfrac{\mathrm dy}{\mathrm dx} = 4\sinh\left(4\tanh^{-1}(x)\right) \cosh^2\left(\tanh^{-1}(x)\right)}[/tex]
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
[tex]\cosh\left(\tanh^{-1}(x)\right) = \dfrac1{\sqrt{1-x^2}}[/tex]
and you can instead work on differentiating that; you would end up with
[tex]\dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{16x^3+16x}{(x^2-1)^3}[/tex]