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

1 Answer
Mar 12, 2018

#dy/dx = (y^3e^-x-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)))/(-y^3-2x^2csc((x^2)/(y^2))cot((x^2)/(y^2)))#

Explanation:

#d/dx (csc((x^2)/(y^2)) = e^-x -y)#
First we find the derivative of the left side of the function #d/dx (csc((x^2)/(y^2))# This is a composition of functions. Meaning we apply the Chain Rule. where #f(x) = csc(x)# and #g(x) = (x^2)/(y^2)# where #f(g(x)) = csc((x^2)/(y^2))#

#d/dx f(x) = csc(x) = -csc(x)cot(x)# we must substitute of #x = g(x)# , so #d/dx csc(g(x)) = -csc((x^2)/(y^2))cot((x^2)/(y^2))#

#d/dx g(x) = (2xy^2-2yx^2dy/dx)/(y^4)# apply the quotient rule #(f'g -fg')/(g)^2# where #f = x^2# and #g = y^2#

Furthermore, #d/dx f(g(x)) = (-csc((x^2)/(y^2))cot((x^2)/(y^2)))(2xy^2-2yx^2dy/dx)/(y^4)# let us simplify this function now as to make our lives easier in the following minutes.
We multiple #(-csc((x^2)/(y^2))cot((x^2)/(y^2)))# by both term #2xy^2# and term #-2yx^2dy/dx# resulting in a function similarly looking to that of

#(2yx^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))-2xy^2csc((x^2)/(y^2))cot((x^2)/(y^2)))/(y^4)# we see that the top function has a like term of #y# , so
#(2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)))/(y^3)#

We are not finished we must now find the derivative of the right side of the function. #d/dx e^-x - y# substituting #u = -x# #d/dx e^u = e^-x#
#d/dx -x = -1#
hence #d/dx e^u = -e^-x#
#d/dx -y = -dy/dx#

Therefore #d/dx e^-x -y = -e^-x-dy/dx#

Furthermore, # (2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)))/(y^3) = -e^-x-dy/dx#

From here we must move #dy/dx# to one side of our function.
Meaning first multiple the left side of the function by #y^3# moving this term to the right. resulting in,
#(2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))-2xycsc((x^2)/(y^2))cot((x^2)/(y^2))) = -y^3e^-x-y^3dy/dx#

From here let us subtract #2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))# from both sides because it has the term #dy/dx# resulting in,
#-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)) = -y^3e^-x-y^3dy/dx-2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))#

Now let us remove any terms on the right side that aren't #dy/dx# , so
#y^3e^-x-2xycsc((x^2)/(y^2))cot((x^2)/(y^2))= -y^3dy/dx-2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))# From here factor out #dy/dx# from the right of our equation #y^3e^-x-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)) = dy/dx(-y^3-2x^2csc((x^2)/(y^2))cot((x^2)/(y^2)))# now divide both sides by our factored expression resulting in our final answer.
#dy/dx = (y^3e^-x-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)))/(-y^3-2x^2csc((x^2)/(y^2))cot((x^2)/(y^2)))#