Question

Find the slope of the graph of the relation 2x^3 − 3xy + y^3 = −1 at the point (2, −3).

Answer 1

You could find the derivative of this curve by implicit differentiation.

\[\frac{ d }{ dx }(2x^{3}-3(xy)) = \frac{ d }{ dx }(-1)\]

Answer 2

\[6x^{2}-3\frac{ d }{ dx}(xy) = 0 \]

\[u = x; v = y, du = 1, dv = y' \]

\[6x^{2}-3(y+x \frac{ dy }{ dx }) = 0\]

Answer 3

further manipulation of this and insetting the x and y values will get you the slope. i'll stop here.

Answer 4

But what about the y^3 @Photon336

Answer 5

I missed that point sorry \[6x^{2}-3(y+x \frac{ dy }{ dx })+3y^{2}(\frac{ dy }{ dx }) = 0\]

Answer 6

what you can do is group terms with dy together

Answer 7

\[6x^{2}-3y+(x+3y^{2})(\frac{ dy }{ dx }) = 0\]

Answer 8

\[3y^{2}-6x^{2} = (x+3y^{2})(\frac{ dy }{ dx })\]

now what you can do is solve for dy/dx and plug in your x and y values. originally given.

Answer 9

\[\frac{ 3y^{2}-6x^{2} }{ x+3y^{2} } = \frac{ dy }{ dx }\]

Answer 10

That makes sense, thank you!