Implicitly differentiate xy+x^2.y^2=4

Find dy/dx.

Hi stuff you,

Begin with:
xy + x²y² = 4.

Recall the product rule for differentiation:
d[fg] = fdg + gdf.

Apply the product rule with f = x and g = y:
d[xy] = xdy + ydx.

Apply the product rule with f = x² and g = y²:
d[x²y²] = x²(2ydy) + y²(2xdx)
= 2x²ydy + 2xy²dx.

Now, the differential of any constant is 0, so:
d[4] = 0.

Differentiating your entire expression, substitute in the values:
d[xy] + d[x²y²] = d[4].
xdy + ydx + 2x²ydy + 2xy²dx = 0.

Factor the differentials on the left-hand side:
dy(x + 2x²y) + dx(y + 2xy²) = 0.

Subtract one from the other:
dy (x + 2x²y) = -dx (y + 2xy²).

Divide by dx:
dy/dx (x + 2x²y) = - (y + 2xy²).

Divide by (x + 2x²y):
dy/dx = - (y + 2xy²) / (x + 2x²y).

You can also factor y from the numerator and x from the denominator if you wish, to get:
dy/dx = - (y / x) ((1 + 2xy) / (1 + 2xy)) = - (y / x) (1) = - (y / x).

Hence,
dy/dx = - y / x.

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Solution:
dy/dx = - y / x.
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Further, you can actually solve this system completely to get:
y = (±√(17) - 1) / (2x).

Implicit Differentiation - UC Davis Math: Welcome
Find y' = dy/dx for x 3 + y 3 = 4 . Click HERE to see a detailed solution to problem 1. PROBLEM 2 : Assume that y is a function of x. Find y' = dy/dx for (x-y) 2 = x + y - 1 .
www.math.ucdavis.edu/~kouba/CalcOneDIR… - Cached

Differentiating implicitly:

xy' + y + 2(xy)*(xy' + y) = 0

now solve for y':

xy' + y + 2x²yy' + 2xy² = 0

y' = (-2xy² - y)/(x + 2x²y)

Assuming the period between x^2.y^2 was supposed to be a division ( / ) sign.

dy/dx = -y^5/[(xy^4+2xy(y-x)]