0
$\begingroup$

I am practicing on the following implicit differentiation problem:

$x^3-xy+y^2=4$

I am stuck at this point, I get:

$3x^2-y+y'x+y'2y$

while derivative-calculator.net gets:

$3x^2−y'x+2y'y−y$

My issue is when I solve for y' I end up with the signs mixed up and I am not sure why.

my incorrect answer:

$y'= \frac{-3x^2+y}{x+2y}\ $

2 Answers 2

0

You have $x^3-xy+y^2=4$, then you have to differentiate. (I broke them into parts)

$\rightarrow \frac{d}{dx}(x^3-xy+y^2=4)$

$\rightarrow \frac{d}{dx}(x^3)=3x^2$

$\rightarrow \frac{d}{dx}(-xy)= -y-xy'$ (Product Rule).

$\rightarrow \frac{d}{dx}(y^2)=2yy'$, and $\frac{d}{dx}(4)=0.$

Then you have $3x^2-y-xy'+2yy'=0$, solve for y'

$\rightarrow y'= \frac{y-3x^2}{2y-x}$.

  • 1
    on the third step, $\frac{d}{dx}(-xy)$ shouldn't that be the product rule? I don't see how the chain rule applies there. ($\frac{d}{dx}x * y + \frac{d}{dx}y * x$ – 2017-02-02
  • 0
    Yeah my bad it was the product rule, but do you understand how they got the -y'x? – 2017-02-02
  • 2
    yes, thanks! I had to write it out as $(-1)((1)(y)+(y')(x)) = -y-xy'$ – 2017-02-02
0

You should be using the product rule $(uv)'=u'v+uv'$ to obtain: $$3x^2-\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0$$ $$3x^2-y-x\frac{dy}{dx}+2y\frac{dy}{dx}=0$$ Now, you can gather all the $\frac{dy}{dx}$ terms to obtain: $$3x^2-y=-2y\frac{dy}{dx}+x\frac{dy}{dx}$$ Now, factor the $\frac{dy}{dx}$ terms, and obtain an explicit expression for $\frac{dy}{dx}$. This should give you the correct answer.