Question

Where does the x^2y’ term come from in 2xy + x^2y’ + 3y^2y’ + 2x = 0? By the way, 2xy + x^2y’ + 3y^2y’ + 2x = 0 is achieved after using implicit differentiation on x^2y + y^3 + x^2 = 8.

Answer 1

\[x^2y+y^3+x^2=8\] take differentiate both sides
the first term is = \(2xy + x^2 y'\) (this is a product, right?)
second term is = \(3y^2y'\)
the third one is = 2x
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\(2xy+x^2y'+3y^2y'+2x=0\)

Answer 2

does it make sense??

Answer 3

\[(x^2\color{red}{y})'= (x^2)'\color{red}{y}+x^2\color{red}{(y)'}=2xy+x^2y' \]
Is it not clear enough??

Answer 4

@charlotte123

Answer 5

@abb0t ;-;

Answer 6

T_T