Find dy/dx by implicit differentiation: tan(x-y) = y/(2+x^2)

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The value of `dy/dx` has to be determined given that `tan(x-y) = y/(2+x^2)`

Using implicit differentiation.

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

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

=> `dy/dx = ((2xy)/(2 +x^2)^2 +...

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The value of `dy/dx` has to be determined given that `tan(x-y) = y/(2+x^2)`

Using implicit differentiation.

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

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

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

The derivative` dy/dx = (2xy + sec^2(x - y)*(2 + x^2)^2)/(2 + x^2 + sec^2(x - y)*(2 + x^2))`

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