Question

find y' by implicit differentiation

Answer 1

\[2x^3+x^2y-xy^3 = 2\]

Answer 2

you know d(2x^3)/dx=?

Answer 3

So I figure you find the derivative normally, so you get \[2(3)x^2+2x*y \frac{ dy }{ dx }-3y^2*\frac{ dy }{ ?dx} =0\]

Answer 4

please excuse the question mark

Answer 5

is that on the right track?

Answer 6

First term is right
second term and third term need a little work

Answer 7

and the right hand side of the equation is correct of course

Answer 8

it looks like you are not using product rule

Answer 9

even know xy is a product
or x^2y is a product
or xy^3

Answer 10

Can you show the product rule for second and third term?

Answer 11

\[\frac{d}{dx}(fg)=g \frac{df}{dx}+f \frac{d g}{dx}\]

Answer 12

\[\frac{d}{dx}(x^ny^m)=y^m \frac{d}{dx}x^n+x^n \frac{d}{dx}y^m \\=y^m \cdot nx^{n-1}+x^n \cdot m y^{m-1} y'\]

Answer 13

\[\text{ notice we needed the chain rule on } \frac{d}{dx}y^m=my^{m-1}y' \]

Answer 14

okay, I think I am starting to understand it.