Question

Find dy/dx by implicit differentiation: x^2y^2 - xy =4

Answer 1

Answer:

dy/dx = -y/x

Step-by-step explanation:

(x²)(2yy') + y²(2x) - [xy' + y] = 0

2x²yy' - xy' = y - 2xy²

y'(2x²y - x) = y - 2xy²

dy/dx = [y(1-2xy)]/[x(2xy-1)]

dy/dx = -y/x

Answer 2

Answer:

$y'=\frac{-y}{x}$

Step-by-step explanation:

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

We will need to differentiate both sides.

Keep in mind the following:

* $\frac{d}{dx}x^2=2x$ by power rule ($(x^n)'=nx^{n-1}$)

* $\frac{d}{dx}y^2=2yy'$ by power rule and chain rule.

* $\frac{d}{dx}x=\frac{dx}{dx}=1$

* $\frac{d}{dx}y=\frac{dy}{dx}=y'$

* $\frac{d}{dx}(xy)=x\frac{d}{dx}y+y\frac{d}{dx}x=x\frac{dy}{dx}+y\frac{dx}{dx}$

  $=x\frac{dy}{dx}+y(1)=x\frac{dy}{dx}+y=xy'+y$ by product rule and some      already mentioned things. This is by product rule ($(uv)'=uv'+vu'$.

* $\frac{d}{dx}(x^2y^2)=x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)$

 $=x^2(2y)y'+y^2(2x)$ by product rules and already mentioned things above.

$=x^22yy'+2xy^2$.

Let's put it all together by differentiating both sides:

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

$(x^2y^2-xy)'=(4)'$

$(x^2y^2)'-(xy)'=0$ I used the constant rule on the right hand side.

I also used difference rule. That is (f-g)'=f'-g'.

Now let's apply those mentioned things:

$(x^22yy'+2xy^2)-(xy'+y)=0$

Distribute:

$x^22yy'+2xy^2-xy'-y=0$

Put $y'$ terms together:

$x^22yy'-xy'+2xy^2-y=0$

Factor the $y'$ out of the terms containing the $y'$:

$y'(x^22y-x)+2xy^2-y=0$

Let's isolate the term containing the $y'$.

We will do this by adding $y$ on both sides and subtracting $2xy^2$ on both sides:

$y'(x^22y-x)=y-2xy^2$

Now divide both sides the thing being multiplied by $y'$.

$y'=\frac{y-2xy^2}{x^22y-x}$

I really don't like the coefficient of $x^2y$ not being in front so I'm going to rearrange that part using the commutative property of multiplication.

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

Let's see if this can be simplified.

I'm going to factor out what I can on both top and bottom.

The top terms contain a factor of $y$.

The bottom terms contain a factor of $x$.

$y'=\frac{y(1-2xy)}{x(2xy-1)}$

We see that the other factor in top and bottom are opposite factors. I'm talking about the $1-2xy$ and the $2xy-1$.

So if we factor out -1 on top we get:

$y'=\frac{-y(2xy-1)}{x(2xy-1)}$

Now we can cancel the common factor across the division there:

$y'=\frac{-y}{x}$