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Find the equation of the tangent line at the given value of x on the curve.

y^3+xy^2- 4=x+4y^2; x=4

User MattCochrane
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1 Answer

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14 votes

Differentiate both sides of

y³ + xy² - 4 = x + 4y²

with respect to x.

Using the power, product, and chain rules,

d/dx [y³] = 3y² dy/dx

d/dx [xy²] = d/dx [x] y² + x d/dx [y²] = y² + 2xy dy/dx

d/dx [4y²] = 8y dy/dx

Putting everything together, we have

3y² dy/dx + y² + 2xy dy/dx = 1 + 8y dy/dx

Solve for dy/dx :

(3y² + 2xy - 8y) dy/dx = 1 - y²

dy/dx = (1 - y²) / (3y² + 2xy - 8y)

From the given equation, when x = 4, we have

y³ + 4y² - 4 = 4 + 4y²

y³ = 8

y = 2

so that the curve passes through the point (4, 2).

When x = 4 and y = 2, the derivative has a value of

dy/dx (4, 2) = (1 - 2²) / (3•2² + 2•4•2 - 8•2) = -1/4

and this is the slope of the tangent line.

Use the point-slop formula to get the line's equation:

y - 2 = -1/4 (x - 4)

y - 2 = -1/4 x + 1

y = -1/4 x + 3

User Kporter
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