Question

What is the derivative of `x = tan (x+y)`?

Answer 1

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

Explanation:

When we differentiate `y` wrt `x` we get `dy/dx`.

However, we only differentiate explicit functions of `y` wrt `x`. But if we apply the chain rule we can differentiate an implicit function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

Example:

`d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx`

When this is done in situ it is known as implicit differentiation.

Now, we have:

`x = tan(x+y)`

Implicitly differentiating wrt `x` (applying product rule):

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

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

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

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

`:. dy/dx = -sin^2(x+y)`

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) = tan(x+y)-x`; Then;

`(partial F)/(partial x) = sec^2(x+y)-1`

`(partial F)/(partial y) = sec^2(x+y)`

And so:

`dy/dx = -(sec^2(x+y)-1)/(sec^2(x+y))`
`" " = -(1-1/(sec^2(x+y)))`
`" " = -(1-cos^2(x-y))`
`" " = -sin^2(x+y)`, as before