Question

Find dy/dx in terms of x & y.

x^3+y^3=3xy^2

How do I even begin?

Answer 1

Implicit differentiation do you know it?

Answer 2

\[y' = \frac{ dy }{ dx }\]

Answer 3

So I'll start you off, \[3x^2 + 3y^2 \frac{ dy }{ dx } = ...\] now apply the product rule

Answer 4

No, I did not know that. Thank you! Should I have 18xy + 9y^2 after applying the product rule?

Answer 5

Sorry my OS crashed

Answer 6

Applying the product rule on the left side we should have \[3y^2+6xy \frac{ dy }{ dx }\]

Answer 7

right side*

Answer 8

So we have \[3x^2+3y^2\frac{ dy }{ dx }=3y^2+6xy \frac{ dy }{ dx }\] now solve for dy/dx

Answer 9

Oh okay that makes sense! Thank you so much!! I got -[(x)^2]/(2xy)

Answer 10

Mhm can you show me how you got that?

Answer 11

\[\frac{ dy }{ dx } 3y ^{2} - 6xy = 3y ^{2}-3x ^{2}\]
Then I moved the things from the left and created a fraction
\[\frac{ 3y ^{2}-3x ^{2} }{ 3y ^{2}-6xy }\]
After that, I simplified

-3x^2/-6xy to -x^2/2xy

Answer 12

\[\frac{ dy }{ dx } = \frac{ x^2-y^2 }{ 2xy-y^2 }\] you get this yeah?

Answer 13

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

Answer 14

\[x^2-y^2 = (2xy-y^2)\frac{ dy }{ dx } \implies \frac{ dy }{ dx } = \frac{ x^2-y^2 }{ 2xy-y^2 }\]

Answer 15

I see what I did wrong. I assumed that you could just cancel the 3y^2/3y^2. I apologize and thank you so much for your time!

Answer 16

Yw :)