Question

Derivatives of Trigonometric, Logarithmic, and Exponential Functions. Implicit Relations. How do I differentiate this implicit function ? Answer: sin(x-y) / sin(x-y) - 1

y = `cos(x-y)`

Answer 1

Given: `y = cos(x-y)`

Differentiate both sides. The left side is trivial:

`dy/dx = (d(cos(x-y)))/dx`

The right side requires the chain rule:

`(d(cos(x-y)))/dx = -sin(x-y)(d(x-y))/dx`

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

Returning to the equation, we substitute into the right side:

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

Distribute the `-sin(x-y)`:

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

Move all of the terms containing `dy/dx` to the left:

`dy/dx - sin(x-y)dy/dx = -sin(x-y)`

factor out `dy/dx`

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

Divide both sides by `(1-sin(x-y))`

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

Multiply the right side by 1 in the form of `(-1)/-1`

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