Question

Find the equation of the tangent at the point (1.1) for the function Y given in the equation

Answer 1

If y = y(x), then the slope of the tangent line to (1, 1) is equal to the value of the derivative dy/dx when x = 1 and y = 1.

Compute the derivative using implicit differentiation:

d/dx [xy ^2 + y] = d/dx [2x]

d/dx [xy ^2] + d/dx [y] = 2 d/dx [x]

(x d/dx [y ^2] + d/dx [x] y ^2) + dy/dx = 2

2xy dy/dx + y ^2 + dy/dx = 2

(2xy + 1) dy/dx = 2 - y ^2

dy/dx = (2 - y ^2) / (2xy + 1)

Plug in x = 1 and y = 1 :

slope = dy/dx = (2 - 1^2) / (2*1*1 + 1) = 1/3

Now use the point-slope formula to get the equation of the line:

y - 1 = 1/3 (x - 1)

y = x/3 + 2/3