How do you use implicit differentiation to find #(dy)/(dx)# given #4x^2=2y^3+4y#?

2 Answers
Nov 27, 2016

I found:
#(dy)/(dx)=(4x)/(3y^2+2)#

Explanation:

We differentiate as usual BUT remembering to include the term #(dy)/(dx)# when we differentiate any #y#; so we get:
#8x=6y^2(dy)/(dx)+4(dy)/(dx)#
Collect #(dy)/(dx)#:
#(dy)/(dx)=(8x)/(6y^2+4)=(8x)/(2(3y^2+2))=(4x)/(3y^2+2)#

Nov 27, 2016

Please see the explanation.

Explanation:

For the x term, you can just use the power rule:

#(d(4x^2))/dx = 8x#

For the y terms, you use the power rule but then multiply by #dy/dx#

#(d(2y^3 + 4y))/dx = (6y^2 + 4)dy/dx#

NOTE: You are really using the chain rule along with the power rule.

Put the equation back together:

#8x = (6y^2 + 4)dy/dx#

Solve for #dy/dx#

#dy/dx = (8x)/(6y^2 + 4)#

Remove #2/2#:

#dy/dx = (4x)/(3y^2 + 2)#

Done.