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Suppose y is a differentiable function of x that satisfies the equation (xy³ + 2x²y = 3). Find(y' ) at the point (1,1).

User Daniaal
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Final answer:

To find y' at the point (1,1), differentiate the equation (xy³ + 2x²y = 3) implicitly with respect to x. Substitute x=1 and y=1, then solve for dy/dx to get the value of y' at the point (1,1).

Step-by-step explanation:

To find y' at the point (1,1), we need to differentiate the equation (xy³ + 2x²y = 3) implicitly with respect to x. Let's differentiate both sides of the equation:

d/dx(xy³) + d/dx(2x²y) = d/dx(3)

Using the product rule and chain rule, we get:
y³ + 3xy²(dy/dx) + 4xy(dy/dx) = 0

We can simplify this to:
3xy²(dy/dx) + 4xy(dy/dx) = -y³

Now, let's substitute x=1 and y=1, since we want to find y' at the point (1,1):
3(1)(1²)(dy/dx) + 4(1)(1)(dy/dx) = -(1)³

Simplifying further, we have:
3(dy/dx) + 4(dy/dx) = -1

Combining like terms:
7(dy/dx) = -1

Finally, solving for dy/dx:
dy/dx = -1/7

User Timothykc
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