Chain rule time here!

But first let's just remember that #tan^-1(x) = arctan(x)#.

Rewriting your expression, we get #y = arctan(7x^2-1)^(1/2)#

We can name #v=7x^2-1# and #u=v^(1/2)#. Thus, #y=arctan(u)#.

So, now, by chain rule definition,

#(dy)/(du)(du)/(dv)(dv)/(dx) = (dy)/(dx)#

Going part by part, here:

#(dy)/(du) = (u')/(1+u^2)#

#(du)/(dv)=(1/2)*(1/v^(1/2))#

#(dv)/(dx)=14x#

Now, getting back to the chain rule definition, we have this:

#(u')/(1+u^2)(1/cancel(2))(1/v^(1/2))cancel(14)7x#

After cancelling the #2# and simplifying #14# as #7#, we can now substitute the values of #u# and #v#, as our objective is to get the answer as function of #x#.

#((1/2)*(1/v^(1/2)))/(1+(v^(1/2))^2)*1/(7x^2-1)*7x#

#(1/2(7x^2-1)^(1/2))/(cancel(1)+7x^2cancel(-1))*1/(7x^2-1)*7x#

We can proceed to some more cancellings:

#(1/2(7x^2-1)^(1/2))/(7x^2)*1/(7x^cancel(2)-1)*cancel(7x)#

And, now, rearrange:

#((7x^2-1)^(1/2)/2)/(7x^2(7x-1))=color(green)((7x^2-1)^(1/2))/color(green)(14x^2(7x-1)#