Question

tan(𝑥−𝑦)=𝑦/1+𝑥^2

dy/dx=?

Answer 1

tan(x-y)=y/(1+x^2)

sec^2(x-y)*(1-dy/dx) = [(y+x^2y)(dy/dx)-(y)(2x)] /[(1+x^2)^2]

sec^2(x-y)*(1-dy/dx) = (dy/dx)[1/(1+x^2)] - 2xy/[(1+x^2)^2]

(dy/dx)[sec^2(x-y) + 1/(1+x^2)^2] = sec^2(x-y) + 2xy/[(1+x^2)^2]

dy/dx = [(sec^2(x-y) + 2xy/(1+x^2)^2)] / [sec^2(x-y) + 1/(1+x^2)^2]


Since sec^2(x-y) = 1+tan^2(x-y) = 1+[y^2/(1+x^2)^2], then:

dy/dx = [1+[y^2/(1+x^2)^2] + 2xy/(1+x^2)^2)] / [1+[y^2/(1+x^2)^2] + 1/(1+x^2)^2]

dy/dx = [(1+x^2)^2+y^2+2xy]/[(1+x^2)^2+y^2+1+x^2]

dy/dx = [x^4+2x^2+y^2+2xy+1]/[x^4+3x^2+y^2+2] <-- Final Answer

Answer 2

Assuming the usual carelessness with parentheses, I'll go with
tan(𝑥-𝑦) = 𝑦/(1+𝑥^2)
sec^2(x-y) (1 - y') = 1/(1+x^2) y' - 2xy/(1+x^2)^2
y' (-sec^2(x-y) - 1/(1+x^2)) = -sec^2(x-y) -2xy/(1+x^2)^2

y' = (sec^2(x-y) + 2xy/(1+x^2)^2) / (sec^2(x-y) + 1/(1+x^2))
Massaging that won't ever make it look any simpler

This is just an exercise in keeping track of the details in the product and chain rules.

Answer 3

I made substitutions in my answer to make it simpler, but I guess both answers are fine. It depends.

Answer 4

I like yours -- gets rid of the trig stuff.
Good call.