Question

Q: The tangent to the curve x^3 + x^2y + 4y = 1 at the point (3,-2) has slope...

Answer 1

use implicit derivative to derive the curve

Answer 2

or u can rewrite it be :
x^3 + x^2y + 4y = 1 or
x^2y + 4y = 1 - x^3
y(x^2 + 4) = 1 - x^3
y = (1 - x^3)/(x^2+4)
use division rule to find dy/dx

Answer 3

if y = u/v then dy/dx = (u'v - uv')/(v^2)

Answer 4

u = 1-x^3 -----> u' = -3x^2
v = x^2 + 4 -----> v' = 2x
just plug of them to the formula above ^
then substitute point x=3 to dy/dx u will get the value of slope

Answer 5

Thank you for the help!

Answer 6

welcome