# Thread: Finding derivative of an xy function containing points

1. ## Finding derivative of an xy function containing points

This question shouldn't be entirely that hard but for some reason when I check with my peers, all of us have completely varying or different answers...

Let f(x) be the continuous function that satisfies the equation $x^4 - 5x^2y^2 + 4y^2 = 0$ and whose graph contains the points $(2,1)$ and $(-2,-2)$. Let [tex]\ell[m/ath] be the line tangent to the graph of $f$ at $x=2$.

A) find the expression for $y '$
B) write an equation for line $\ell$
C) give the coordinates of a point that is on the graph of $f$ but is not on line $\ell$
D) give coordinates of a point that is on line $\ell$ but is not on the graph of $f$

So for PART A, I did this.

$y' = 4x^3 - 10x^2y\frac{dy}{dx} - 10xy^2 + 16y^3\frac{dy}{dx} = 0$

$-10x^2y\frac{dy}{dx} +16y^3\frac{dy}{dx} = -4^3 + 10xy^2$

$\frac{dy}{dx}(16y^3 - 10x^2y) = 10xy^2 - 4x^3$

$\frac{dy}{dx} = \frac{10xy^2 - 4x^3}{16y^3 - 10x^2y} = \frac{5xy^2 - 2x^3}{8y^3 - 5x^2y}$

Then for PART B, I plugged in (2,1) into my dy/dx and I got the slope to be 1/2 and thus my tangent line equation for $\ell$ was $y = \frac{1}{2}x + 0$

Then for PART C, I set the equation to $y(2) = 0$ for $x=2$ and I got that $y = 1, -1, 2, -2$. But I'm not sure what to do with this...

So this is the work I did so far and I'm not sure on how to figure out part C or D of the question... If anyone could correct or check my initial work to see that it's correct or help me with Part C and D, then I would be really grateful! Because with some peers I was checking my work with, they got different derivatives or for part B got slopes like -3/14 or -13/14.

2. Originally Posted by choi_siwon
This question shouldn't be entirely that hard but for some reason when I check with my peers, all of us have completely varying or different answers... Here is the question. (Scanned the question from my worksheet)

So for PART A, I did this.

$y' = 4x^3 - 10x^2y\frac{dy}{dx} - 10xy^2 + 16y^3\frac{dy}{dx} = 0$

$-10x^2y\frac{dy}{dx} +16y^3\frac{dy}{dx} = -4^3 + 10xy^2$

$\frac{dy}{dx}(16y^3 - 10x^2y) = 10xy^2 - 4x^3$

$\frac{dy}{dx} = \frac{10xy^2 - 4x^3}{16y^3 - 10x^2y} = \frac{5xy^2 - 2x^3}{8y^3 - 5x^2y}$

Then for PART B, I plugged in (2,1) into my dy/dx and I got the slope to be 1/2 and thus my tangent line equation for $\ell$ was $y = \frac{1}{2}x + 0$

Then for PART C, I set the equation to $y(2) = 0$ for $x=2$ and I got that $y = 1, -1, 2, -2$. But I'm not sure what to do with this...

So this is the work I did so far and I'm not sure on how to figure out part C or D of the question... If anyone could correct or check my initial work to see that it's correct or help me with Part C and D, then I would be really grateful! Because with some peers I was checking my work with, they got different derivatives or for part B got slopes like -3/14 or -13/14.
Where is the original problem?

3. You should spend some time trying to reduce this before you do any calculus.

$x^4-5x^2y^2+4y^4=(x^2-4y^2)(x^2-y^2)=(x-2y)(x+2y)(x-y)(x+y)$

4. Original problem is:

Let f(x) be the continuous function that satisfies the equation $x^4 - 5x^2y^2 + 4y^2 = 0$ and whose graph contains the points $(2,1)$ and $(-2,-2)$. Let [tex]\ell[m/ath] be the line tangent to the graph of $f$ at $x=2$.

A) find the expression for $y '$
B) write an equation for line $\ell$
C) give the coordinates of a point that is on the graph of $f$ but is not on line $\ell$
D) give coordinates of a point that is on line $\ell$ but is not on the graph of $f$

Sorry if the image didn't show.

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matheagle: I thought it would be okay just find the derivative using the power rule, while using product rule uv'+vu' for the $- 5x^2y^2$. Since I separated it as

$- 5x^2y^2 = (-5x^2)(y^2)$
$uv' + vu' = (-5x^2)(2y\frac{dy}{dx}) + (y^2)(-10x) = -10x^2y\frac{dy}{dx} - 10xy^2$ <-- thus how I got middle part

I'm not sure how to utilize the reduced form of the equation to help me on this problem, could you please explain it a little to me? (Sorry, I'm not so good at calculus . )