How do you implicitly differentiate #xy^2-x/y=-1#?

Question

1206 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T20:21:24

#(dy)/(dx)=(1-y^3)/(3xy^2+1)#

Firstly multiply everything by #y# to get rid of the awkward denominator.

#xy^3-x=-y#

now differentiate both sides wrt #x# remembering that any #f(y)# will be differentiated wrt #y# then multiplied by #(dy)/(dx)#

Also the first term will need the use of the product rule.

#d/(dx)(xy^3)-d/(dx)(x)=d/(dx)(-y)#

#y^3+x3y^2(dy)/(dx)-1=-(dy)/(dx)#

now rearrange for#" " (dy)/(dx)#

#x3y^2(dy)/(dx)+(dy)/(dx)=1-y^3#

#(dy)/(dx)[3xy^2+1]=1-y^3#

#(dy)/(dx)=(1-y^3)/(3xy^2+1)#