How do you implicitly differentiate #2=e^ysinx-x^2y^3 #?

1 Answer
Feb 29, 2016

I found: #(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-3x^(2)y^2)#

Explanation:

You need to consider #y# as a function of #x# and so differentiate it accordingly. For example:
if you have #y^2# differentiating you'll get #2y(dy)/(dx)# to consider the fact that #y# represents a function of something!
in our case:
#0=e^y(dy)/(dx)sin(x)+e^ycos(x)-2xy^3-x^(2)3y^2(dy)/(dx)#
Here I used the Product Rule and every time I differentiated #y# I needed to include #(dy)/(dx)#:

collect #(dy)/(dx)#:
#(dy)/(dx)[e^ysin(x)-x^(2)3y^2]=2xy^3-e^ycos(x)#
and:
#(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-x^(2)3y^2)#