Question

How do you find `dy/dx` by implicit differentiation given `x=siny+cosy`?

Answer 1

`dy/dx = 1/(cos y -sin y)`

Explanation:

We have:

`x = sin y + cos y`

Method 1 - Use the Chain Rule

Differentiate wrt `y`;

`dx/dy = cos y -sin y`

By the chain rule; `dx/dy = 1/(dy/dx)`; so:

`dy/dx = 1/(cos y -sin y)`

Method 2 - Implicit Differentiation

Differentiate wrt `x`;

`\ \ \ \ \ \ \ \ 1 = d/dx(sin y + cos y)`
`:. \ \ \ 1 = dy/dx*d/dy(sin y + cos y)`
`:. \ \ \ 1 = dy/dx*(cos y -sin y)`
`:. dy/dx = 1/(cos y -sin y)`