How do you implicitly differentiate #2=e^(x-y)/tan(y)#?

1 Answer
May 4, 2017

Please see the explanation.

Explanation:

Given: Implicitly differentiate #2=e^(x-y)/tan(y)#

Change #1/tan(y) to cot(y)#:

#2=e^(x-y)cot(y)#

Differentiate both sides:

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

The left side is 0:

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

Use the product rule on the right side:

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

Substitute #-csc^2(y)dy/dx# for #(d(cot(y)))/dx#

#0= (d(e^(x-y)))/dxcot(y)+ e^(x-y)(-csc^2(y)dy/dx)#

Write #e^(x-y)# as #e^xe^-y#:

#0= (d(e^xe^-y))/dxcot(y)+ e^(x-y)(-csc^2(y)dy/dx)#

Use the product rule:

#0= (e^xe^-y-e^xe^-ydy/dx)cot(y)+ e^(x-y)(-csc^2(y)dy/dx)#

Return to the #e^(x-y)# form:

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

Use the distributive property:

#0= e^(x-y)cot(y)-e^(x-y)cot(y)dy/dx- e^(x-y)csc^2(y)dy/dx#

Move the terms containing #dy/dx# to the left:

#e^(x-y)cot(y)dy/dx +e^(x-y)csc^2(y)dy/dx= e^(x-y)cot(y)#

Factor out #dy/dx#:

#(e^(x-y)cot(y) +e^(x-y)csc^2(y))dy/dx= e^(x-y)cot(y)#

#e^(x-y)# is on both sides so it cancels:

#(cot(y) +csc^2(y))dy/dx= cot(y)#

Divide by sides by #cot(y) +csc^2(y)#

#dy/dx= cot(y)/(cot(y) +csc^2(y))#