How do you implicitly differentiate #4y^2= x^3y+y-x^2y #?

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Answer 1 · 2016-07-21T19:16:07

#(dy)/(dx) = (3x^2y - 2xy)/(8y + x^2 - x^3 - 1)#

Let's walk through this term by term.

First term, y is a function of x, using the fact that #d/(dx)(y) = (dy)/(dx)# and the chain rule we obtain that

#d/(dx)(4y^2) = 8y(dy)/(dx)#

Second term, we need to use the product rule:

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

Third:

#d/(dx)(y) = (dy)/(dx)#

Fourth:

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

Combining these gives:

#8y(dy)/(dx) = 3x^2y + x^3(dy)/(dx) + (dy)/(dx) -2xy - x^2(dy)/(dx)#

Gather like terms:

#(dy)/(dx)(8y + x^2 - x^3 - 1) = 3x^2y - 2xy#

#(dy)/(dx) = (3x^2y - 2xy)/(8y + x^2 - x^3 - 1)#