Find the derivative of x²y+y²=x³ using implicit differentiation and find the equation of line tangent at 1,-3

Question

I just want to double check my work. I came up with y=1/2x-3, but I'm not so sure about the y-intercept primarily because all of the problems I worked with usually had 0 as the point in x, so obviously I knew what the intercept was. Please help. Thanks!

Answer 1

yx^2 + y^2 = x^3 ----> (2xy + y' * x^2) +2y*y' = 3x^2

y' * x^2 +2y*y' = 3x^2 - 2xy

y'(x^2 + 2y) = 3x^2 - 2xy

y' = (3x^2 - 2xy) / (x^2 + 2y)
=(3(1)^2 - 2(1)(-3)) / ((1)^2 + 2(-3)) = -9 / 5

You don't use y-intercpt form, use point-slope form:

y - y1 = m(x - x1)

y + 3 = (-9/5)(x - 1)
y = (-9/5)x - (6/5)
5y = -(9x + 6)

Answer 2

2xy + x²y' + 2yy' = 3x²
x²y' + 2yy' = 3x² - 2xy
y'(x² + 2y) = 3x² - 2xy
y' = (3x² - 2xy)/(x² + 2y)

at (1,-3), y' = (3+6)/(1-6) = -9/5

So the slope of the tangent line is -9/5 and the point of tangency is (1,-3). Initially using point slope equation yields:

y+3 = (-9/5) (x-1) which is
y = (-9/5)x -6/5

Answer 3

x²y+y²=x³
2xy + x^2y' +2yy' =3x^2
x^2y' +2yy' = 3x^2-2xy
y'(x^2+2y) = 3x^2-2xy
y' = (3x^2-2xy)/((x^2+2y)
At (1,-3), y'=(3+6)/(1 -6) = - 9/5
y = -9x/5 +b
-3 = -9/5 +b
b = 9/5-3 = -6/5
So y= -9x/5 -6/5