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)#