I'm assuming you want to find (dy)/(dx)dydx. For this we first need an expression for yy in terms of xx. We note that this problem has various solutions, since tan(x)tan(x) is a periodic functions, tan(x-y)=xtan(x−y)=x will have multiple solutions. However, since we know the period of the tangent function (piπ), we can do the following: x-y=tan^(-1)x+npix−y=tan−1x+nπ, where tan^(-1)tan−1 is the inverse function of the tangent giving values between -pi/2−π2 and pi/2π2 and the factor npinπ has been added to account for the periodicity of the tangent.
This gives us y=x-tan^(-1)x-npiy=x−tan−1x−nπ, therefore (dy)/(dx)=1-d/(dx)tan^(-1)xdydx=1−ddxtan−1x, note that the factor npinπ has disappeared. Now we need to find d/(dx)tan^(-1)xddxtan−1x. This is quite tricky, but doable using the reverse function theorem.
Setting u=tan^(-1)xu=tan−1x, we have x=tanu=sinu/cosux=tanu=sinucosu, so (dx)/(du)=(cos^2u+sin^2u)/cos^2u=1/cos^2udxdu=cos2u+sin2ucos2u=1cos2u, using the quotient rule and some trigonometric identities. Using the inverse function theorem (which states that if (dx)/(du)dxdu is continuous and non-zero, we have (du)/(dx)=1/((dx)/(du))dudx=1dxdu), we have (du)/(dx)=cos^2ududx=cos2u. Now we need to express cos^2ucos2u in terms of x.
To do this, we use some trigonometry. Given a right triangle with sides a,b,ca,b,c where cc is the hypotenuse and a,ba,b connected to the right angle. If uu is the angle where side cc intersects side aa, we have x=tanu=b/ax=tanu=ba. With the symbols a,b,ca,b,c in the equations we denote de length of these edges. cosu=a/ccosu=ac and using Pythagoras theorem, we find c=sqrt(a^2+b^2)=asqrt(1+(b/a)^2)=asqrt(1+x^2)c=√a2+b2=a√1+(ba)2=a√1+x2. This gives cosu=1/sqrt(1+x^2)cosu=1√1+x2, so (du)/(dx)=1/(1+x^2)dudx=11+x2.
Since u=tan^(-1)xu=tan−1x, we can substitute this into our equation for (dy)/(dx)dydx and find (dy)/(dx)=1-1/(1+x^2)=x^2/(1+x^2)dydx=1−11+x2=x21+x2.
