What is the derivative of $tan(x-y)=y/(1+x^2)$ ?

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Answer 1 · 2018-05-07T14:31:25

$dy/dx = (x^4+2x^2 + y^2 + 2xy+1)/(x^4+3x^2 + y^2 +2)$

Differentiate both sides of the equality with respect to $x$ :

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

solve for $dy/dx$ :

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

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

Note now that:

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

so:

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

and multiplying numerator and denominator by $(1+x^2)^2$ :

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

$dy/dx = (x^4+2x^2 + y^2 + 2xy+1)/(x^4+3x^2 + y^2 +2)$

What is the derivative of tan(x-y)=y/(1+x^2) tan(x-y)=y/(1+x^2)?