- #1
James2
- 35
- 0
1. Given that [itex]y^{2}-2xy+x^{3}=0[/itex], find [itex]\frac{dy}{dx}[/itex]
2. (no relevant equations other than the problem statement)
3. So, I solved it like this,
[itex]\frac{dy}{dx}y^{2}-2xy+x^{3}=0[/itex]
[itex]2y\frac{dy}{dx}-2+3x^{2}=0[/itex]
Solving for dy/dx I got...
[itex]\frac{dy}{dx}=\frac{-3x^{2}+2}{2y}[/itex]
However, wolfram alpha on my phone when I checked it said that the derivative of -2xy was -2y...? What I am asking is, if there are two variables in one term, the derivative is different?
2. (no relevant equations other than the problem statement)
3. So, I solved it like this,
[itex]\frac{dy}{dx}y^{2}-2xy+x^{3}=0[/itex]
[itex]2y\frac{dy}{dx}-2+3x^{2}=0[/itex]
Solving for dy/dx I got...
[itex]\frac{dy}{dx}=\frac{-3x^{2}+2}{2y}[/itex]
However, wolfram alpha on my phone when I checked it said that the derivative of -2xy was -2y...? What I am asking is, if there are two variables in one term, the derivative is different?