Question

How do you find `dy/dx` by implicit differentiation of `coty=x-y`?

Answer 1

We start by defining the derivative of `coty`. Recall that `cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx`.

`:.coty = cosy/siny`

`:.(coty)' = (-siny xx siny(dy/dx) - (cosy xx cosy)dy/dx)/(siny)^2`

`:.(coty)' = (-sin^2y(dy/dx) - cos^2y(dy/dx))/sin^2y`

`:.(coty)' = -(sin^2y(dy/dx) + cos^2y(dy/dx))/sin^2y`

`:.(coty)' =- (dy/dx)/sin^2y`

`:.(coty)' = -dy/dxcsc^2y`

Now, the rest of the relation.

`-csc^2y(dy/dx) = 1 - 1(dy/dx)`

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

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

`dy/dx = 1/(1 - csc^2y)`

From the relation `1+cot^2y=csc^2y` we see that `1-csc^2y=-cot^2y`. We can rewrite the derivative if we wish.

`dy/dx=-1/cot^2y`

If we want, we can use the original function `coty=x-y`.

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

Hopefully this helps!