Question

Use Implicit Differentiation To Find Y'. X² 6xy 12y² = 28 Y' =_____ Find An Equation Of The Tangent Line To The Give Curve At The Point (2, 1).

Answer 1

To find y', use implicit differentiation on the given equation. Then, substitute the values of x=2 and y=1 into y' to find the slope. Finally, use the point-slope form to write the equation of the tangent line.

Step-by-step explanation:

To find y', we can use implicit differentiation. Differentiate both sides of the equation with respect to x. Let's first find the derivative of each term:

d/dx(x²) = 2x

d/dx(6xy) = 6xy' + 6y

d/dx(12y^2) = 24yy'

d/dx(28) = 0

Now, let's simplify and solve for y': 2x + 6xy' + 6y + 24yy' = 0

Combining like terms, we get: 6xy' + 24yy' = -2x - 6y

Factor out y' on the left side: (6x + 24y)y' = -2x - 6y

Divide both sides by (6x + 24y): y' = (-2x - 6y)/(6x + 24y)

Now that we have found y', we can find the equation of the tangent line at the point (2,1). Substitute x=2 and y=1 into y', and calculate the slope of the line:

y' = (-2(2) - 6(1))/(6(2) + 24(1)) = -10/42 = -5/21

Using the point-slope form of a line, the equation of the tangent line is: y - 1 = (-5/21)(x - 2)