Question

How do you implicitly differentiate `tan(x+y) = x`?

Answer 1

`(d(y))/(dx)=(1-sec^2(x+y))(cos^2(x+y))` `...........(i)`
`(d(x))/(dy)=1/((1-sec^2(x+y))(cos^2(x+y)))` `...............(ii)`

Explanation:

`tan(x+y)=x`
Regard `y` as a function of x and differentiate with respect to `x`.
`implies (d(tan(x+y)))/dx=(d(x))/dx`
using chain rule
`implies sec^2(x+y)((d(x))/(dx)+(d(y))/(dx))=1`
`implies sec^2(x+y)(1+(d(y))/(dx))=1`
`implies (d(y))/(dx)=1/sec^2(x+y)`
`implies (d(y))/(dx)=1/sec^2(x+y)-1`
`implies (d(y))/(dx)=(1-sec^2(x+y))/sec^2(x+y)`
`implies (d(y))/(dx)=(1-sec^2(x+y))(cos^2(x+y))` `...........(i)`
`implies (d(x))/(dy)=1/((1-sec^2(x+y))(cos^2(x+y)))` `...............(ii)`

`(i)` represents the derivative of `y` with respect to `x` and `(ii)` represents the derivative of `x` with respect to `y`.