`tan(x - y) = y/(1 + x^2)` Find `(dy/dx)` by implicit differentiation.

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Chapter 3, 3.5 - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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

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

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

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

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

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

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

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