Given y=arctan(3e^2x). Show dy/dx= 3/(5cosh(2x) + 4sinh(2x))

As we cannot immediately differentiate inverse tan we take tan of both sides giving us tan(y)=3e^2x. We then differentiate with respect to x, however this must be done using implicit differentiation on the left hand side as it is a function of y. LHS: d/dx(tan(y)) = sec²(y) * (dy/dx), RHS: d/dx(3e^2x) = 6e^2xN. Now we have: sec²(y) * (dy/dx) = 6e^2x. Which rearranges to: dy/dx = (6e^2x)/sec²(y). Due to trigonometric identities it can be said: sec²(y) = tan²(y) + 1. Subbing into our last equation for dy/dx: dy/dx = (6e^2x)/(tan²(y) + 1). The question stated that: tan(y)=3e^2x , meaning if we square this we get: tan²(y)=9e^4x. So: dy/dx = (6e^2x)/(9e^4x+ 1). Now there are no terms involving trigonometric functions it can be manipulated into the form the final answer wants it in. To start this we must have all exponentials terms on the denominator, as this is where the hyperbolic functions are in the final answer. To do this we times top and bottom by e^-2x giving us: dy/dx = 6/(9e^2x+ e^-2^x). From the answer we can see it must involve hyperbolic functions and using the definitions of them on the denominator we get: 9e^2x+ e^-2^x = a(e^2x+ e^-2^x ) + b(e^2x-e^-2^x ) and we can see that we get simulations equations that state: (a+b)=9, (a-b)=1, Solving for:a= 5, b=4. Subbing this into our equation dy/dx, dy/dx = 6/(5(e^2x+ e^-2^x)+4(e^2x-e^-2^x)). Putting this into the form with hyperbolic functions by dividing top and bottom by 2; dy/dx = 3/(5/2(e^2x+ e^-2^x)+4/2(e^2x-e^-2^x)), which finally give us dy/dx = 6/(5cosh(2x)+4sinh(2x)) as is stated in the proof answer.