Y=5arctanh(cosx)

We need to use the chain rule, here

We have two fucnctions...the "outside' function," arctanh(cosx) 

And the "inside' function," cos x

Let cos x = u  and     u' = -sinx

The derivative of arctanh(u)  is just      1 /( 1 - u^2)

And the derivative of u is just u' 

And we can think of Y' as being the "derivative of the outside function" times the "derivative of the inside function"....so we have...making the substitutions

Y' = 1/[1 - cos^2(x)] * (-sin(x))   = -sin(x)/ [sin^2(x)] = -1/sin(x) =-csc(x)

And we can just "tack" the "5" back on....so we end up with

Y'[5arctanh(cosx)] = -5csc(x)

Hope that helps...!!!