# How do you implicitly differentiate #-1=xycot^2(x-y) #?

##### 1 Answer

# dy/dx = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}#

#### Explanation:

When we differentiate

However, we cannot differentiate a non implicit function of

When this is done in situ it is known as implicit differentiation.

We have:

# -1=xycot^2(x-y) #

Differentiate wrt

# 0 = (x)(y)(d/dx cot^2(x-y)) + (x)(d/dx y)(cot^2(x-y)) + (d/dx x)(y)(cot^2(x-y)) #

# :. 0 = (x)(y)(2cot(x-y)(-csc^2(x-y))(1-dy/dx)) + (x)(dy/dx)(cot^2(x-y)) + (1)(y)(cot^2(x-y)) #

# :. 0 = -2xycot(x-y)csc^2(x-y)(1-dy/dx) + xdy/dxcot^2(x-y) + ycot^2(x-y) #

# :. 2xycot(x-y)csc^2(x-y)dy/dx + xdy/dx(cot^2(x-y) = 2xycot(x-y)csc^2(x-y) - ycot^2(x-y)#

# :. {2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}dy/dx = 2xycot(x-y)csc^2(x-y) - ycot^2(x-y)#

# :. dy/dx = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}#

**Advanced Calculus**

There is another (often faster) approach using partial derivatives. Suppose we cannot find

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let

# (partial F)/(partial x) = (xy)((partial)/(partial x) cot^2(x-y)) + ((partial)/(partial x) xy)(cot^2(x-y)) #

# \ \ \ \ \ \ \ = (xy)(2cot(x-y)(-csc^2(x-y))) + (y)(cot^2(x-y))#

# \ \ \ \ \ \ \ = -2xycot(x-y)csc^2(x-y) + ycot^2(x-y) #

# (partial F)/(partial x) = (xy)((partial)/(partial y) cot^2(x-y)) + ((partial)/(partial y) xy)(cot^2(x-y)) #

# \ \ \ \ \ \ \ = (xy)(2cot(x-y)(-csc^2(x-y)(-1)) + (x)(cot^2(x-y)) #

# \ \ \ \ \ \ \ = 2xycot(x-y)csc^2(x-y) + xcot^2(x-y) #

And so:

# dy/dx = -{-2xycot(x-y)csc^2(x-y) + ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}#

# \ \ \ \ \ = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}#

Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier