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)

1 Answer
May 27, 2017

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)