Question

How do you take the derivative of `tan(x - y) = y/(1 + x^2)`?

Answer 1

Have a look, but check my maths.

Explanation:

In this case you'll need to derive implicitly considering that here `y` represents a function of `x` (a little bit...difficult to work out!) and we derive also `y`:
so if you want, say, to derive `y` you get `1*(dy)/(dx)`

so we get (using the Chain Rule on the left and the Quotient Rule on the right):

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

rearrange to isolate `(dy)/(dx)`:

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

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

Multiply the non-fractions by `cos^2(x-y)/cos^2(x-y)` to get common denominators, cancel out those denominators, then divide by the resultant left-side bracketed contents:

`(dy)/(dx)=[2xycos^2(x-y)+(1+x^2)^2]/[(1+x^2)cos^2(x-y)+(1+x^2)^2]`