Question

Find dy/dx at (0,1) for the graph of x^2+2xy+y^3=1.

Answer 1

Derivative with respect to x:\[\frac{ d(x^{2} + 2xy + y^{3}) }{ dx } = \frac{ d(1) }{ dx }\]\[\frac{ d(x^{2)} }{ dx } + \frac{ d(2xy) }{ dx } + \frac{ d(y^{3}) }{ dx } = 0\]

Answer 2

\[2x + 2y + 2x \times \frac{ dy }{ dx } + 3y^{2} \times \frac{ dy }{ dx } = 0\]Got it @cuzzin ?

Answer 3

use implicit differentiation, if I am correct

Answer 4

then plug the values of x and y to get your answer

Answer 5

At (0,1)
\[2*0 + 2*1 + 2*0*\frac{ dy }{ dx } + 3 * 1^{2} \frac{ dy }{ dx } = 0\]

Answer 6

to get correct answer, make dy/dx the subject