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)#
