Question

Implicit differentiation, am I solving it correctly?

Answer 1

\[x^2-2xy+y^3=10\]

Answer 2

find dy/dx

Answer 3

so\[2x*\frac{ dx }{ dy }-2\frac{dx}{dy}*\frac{dy}{dx}+3y^2*\frac{dy}{dx}=0\]

Answer 4

I want to solve for dy/dx right

Answer 5

if I divide both sides by 2x∗dxdy−2dxdy then it's equal 0

Answer 6

Oh wait I have to use the product rule

Answer 7

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

Answer 8

\[f(x)=x^2−2xy+y^3=10\]\[f'(x)=2x-(2y+2xy')+3y^2\cdot y' = 0\] You see how I did that?

Answer 9

So your setup was correct :) I was just checking with solving it as well.

Answer 10

Whats your next step?

Answer 11

move the 2xdx/dy to the other side?

Answer 12

Yes. \[f'(x):-(2y+2xy')+3y^2\cdot y' = -2x\] Now we want to distribute that negative value so we can gather all the \(y'\) terms.

Answer 13

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

Answer 14

oops the 2x*dx/dy is negative

Answer 15

Mmhmm. \[f'(x):-2y-2xy'+3y^2\cdot y' = -2x\]Now we can send \(-2y\) to the other side as well.

Answer 16

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

Answer 17

I factored the left side.

Answer 18

oops the = sign after dy/dx is a typo

Answer 19

Hmm, let me compare.

Answer 20

\[f'(x):-2y-2xy'+3y^2\cdot y' = -2x\]\[f'(x): -2xy'+3yy'=-2x+2y\]\[f'(x): y'(-2x+3y)=-2x+2y\]\[f'(x): \boxed{\large y' = \frac{-2x+2y}{-2x+3y}}\]

Answer 21

so you don't take the derivative of the other stuff?