How do you find #(dy)/(dx)# given #-3x^2y^2-2y^3+5=5x^2#?

Question

4952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-22T07:19:13

#(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)#

we want

#d/(dx)(-3x^2y^2)-d/(dx)(2y^3)+d/(dx)(5)=d/(dx)(5x^2)#

We will differentiate implicitly.

The first term will also need the product rule

#color(red)(d/(dx)(uv)=v(du)/(dx)+u(dv)/(dx))#

#y^2(-6x)+(-3x^2)2y(dy)/(dx)-6y^2(dy)/(dx)+0=10x#

#-6xy^2-6x^2y(dy)/(dx)-6y^2(dy)/(dx)=10x#

now rearrange for #(dy)/(dx)# and tidy up.

#(dy)/(dx)(-6x^2y-6y^2)=10x+6xy^2#

#(dy)/(dx)=(10x+6xy^2)/(-6y(x^2+y)#

#(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)#