how do I solve this?

Question

find the slope of the tangent line to the curve given by x^2+xy^2+2y=8 at the point (2,1)

Answer 1

The slope of any straight line is given by y=Mx+b , m = slope, b = y-intercept.

We need to find m and b, so let’s fill in the info we already know;

y = mx+b

1=m(2)+b

Now we find slope of the line, m. Since we are finding the slope of a tangent line, we are finding the slope at one point. Meaning, we find instantaneous rate of change, the Derivative.

Find the derivative at (2,1), this is m

(d/dx)(x^2 + xy^2 + 2y) = (d/dx)8

Here, we are doing implicit differentiation. We are taking the d/dx of each side, and we treat y as a variable. Everywhere you take the derivative of y, you must multiply the result by dy/dx, or if you would like, y’.

2x + (x)’(y^2) + (x)(y^2)’ + 2(1)(y’) = 0

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product rule

2x + (1)(y^2) + (x)(2y)(y’) + 2y’ = 0

2x + y^2 + 2xyy’+2y’ = 0

Solve for y’

2xyy’ + 2y’ = -2x - y^2

y’(2xy + 2) = -2x - y^2

y’ = (-2x - y^2)/(2xy + 2)

plug in coordinate

y’ = [-2(2) - (1^2)]/[2(2)(1) + 2] = m

m = -5 /6

We have y, m, and x. Plug in values to find b:

y = Mx + b

2 = (-5/6)(1) + b

b = 17/6

Therefore, the equation of the tangent line at (2,1) is:

y=(-5/6)x+(17/6)

Answer 2