Question

Find dy/dx for the equation x^3-2x^2+4xy^2=20
I was instructed to use the product rule to solve, but Im confused how to actually do that. I know the product rule is f(x)g'(x)+g(x)f'(x) i'm just not really sure how to apply it to the given equation.

Answer 1

sorry i type the equation worn its x^3-2x^2y+4xy^2=20

Answer 2

wrong*

Answer 3

\(\frac{dy}{dx}(x^3-2x^2y+4xy^2)=\frac{dy}{dx}(20)\\\frac{dy}{dx}(x^3)-2\frac{dy}{dx}(x^2y)+4\frac{dy}{dx}(xy^2)=0\)

Answer 4

with me?

Answer 5

hmmm well... 3rd term there, uses the product rule, yes
the others do not

Answer 6

\(\frac{dy}{dx}(x^3-2x^2y+4xy^2)=\frac{dy}{dx}(20)\\\frac{dy}{dx}(x^3)-2\frac{dy}{dx}(x^2y)+4\frac{dy}{dx}(xy^2)=0\\\frac{dy}{dx}x^3-2(\frac{dy}{dx}(x^2)y+\frac{dy}{dx}(y)x^2)+4(\frac{dy}{dx}(x)y^2+\frac{dy}{dx}(y^2)x)=0\)

Answer 7

\(\frac{dy}{dx}(x^3-2x^2y+4xy^2)=\frac{dy}{dx}(20)\\\frac{dy}{dx}(x^3)-2\frac{dy}{dx}(x^2y)+4\frac{dy}{dx}(xy^2)=0\\\frac{dy}{dx}x^3-2[\frac{dy}{dx}(x^2)y+\frac{dy}{dx}(y)x^2]+4[\frac{dy}{dx}(x)y^2+\frac{dy}{dx}(y^2)x]=0\)

Answer 8

i use the product rule inbetween the []

Answer 9

am I talking to my self?

Answer 10

No i'm just trying to go through it and see if I'm picking up on it

Answer 11

word

Answer 12

I know this is a really stupid question but do you have to put the dy/dx in front of each number like you did ?

Answer 13

not really, you could just take the derivative, but I figured this would help you see exactly what we are doing

Answer 14

okay thanks

Answer 15

now since we are doing implicit differentiation we have

\(\frac{dy}{dx}(x^3-2x^2y+4xy^2)=\frac{dy}{dx}(20)\\\frac{dy}{dx}(x^3)-2\frac{dy}{dx}(x^2y)+4\frac{dy}{dx}(xy^2)=0\\\frac{dy}{dx}x^3-2[\frac{dy}{dx}(x^2)y+\frac{dy}{dx}(y)x^2]+4[\frac{dy}{dx}(x)y^2+\frac{dy}{dx}(y^2)x]=0\\
3x^2-2[2xy+\frac{dy}{dx}x^2]+4[y^2+2y\frac{dy}{dx}x]=0\\3x^2-4xy-2\frac{dy}{dx}x^2+4y^2+8yx\frac{dy}{dx}=0\)

Answer 16

now solve for \(\frac{dy}{dx}\)