Differentiating Implicitly

**Hakhengkim** · April 20th 2009, 09:11 AM

Is it possible to differentiate this equation?

(2x^2)(y)-(7y^2)=(8x)-2y-11 to find y'(1)?

**derfleurer** · April 20th 2009, 09:18 AM

2yx^2 - 7y^2 - 8x + 2y = -11

-(d/dx)/(d/dy)

(8 - 4xy) / (2x^2 - 14y + 2)

y'(1) just accounts for one of those variables, however. You need an x and a y coordinate to perform the calculation. For y'(1), do you mean x by that 1? If so, just plug back into the original equation to get the y value accompanying it.

Edit: Missed that 2y', sorry. Fixed

**Hakhengkim** · April 20th 2009, 09:19 AM

Originally Posted by derfleurer

2yx^2 - 7y^2 - 8x + 2y = -11

-(d/dx)/(d/dy)

(8 - 4xy) / (2x^2 - 14y)

y'(1) just accounts for one of those variables, however. You need an x and a y coordinate to perform the calculation.

I was wondering why I couldn't do it either. Thank you.

**Soroban** · April 20th 2009, 10:25 AM

Hello, Hakhengkim!

Did you follow derfleurer's advice and finish the problem?

$2x^2y-7y^2\:=\:8x-2y-11$ . . Find $y'(1)$

Differentiate implicitly . . .

. . $2x\,\frac{dy}{dx} + 4xy - 14y\,\frac{dy}{dx} \:=\:8 - 2\,\frac{dy}{dx} \quad\Rightarrow\quad 2x\,\frac{dy}{dx} - 14y\,\frac{dy}{dx} + 2\,\frac{dy}{dx} \:=\:8 - 4xy$

Factor: . $2(x-7y - 6)\,\frac{dy}{dx} \:=\:4(2-xy) \quad\Rightarrow\quad \frac{dy}{dx} \;=\;\frac{2(2-xy)}{x-7y-6}$

If $x = 1$ , the equation becomes: . $2y - 7y^2 \:=\:8 - 2y - 11 \quad\Rightarrow\quad 7y^2 - 4y - 3 \:=\:0$

. . $(y - 1)(7y+3) \:=\:0 \quad\Rightarrow\quad y \:=\:1,\:-\tfrac{3}{7}$

And we have two points: . $(1,\,1),\;\;\left(1,\,-\tfrac{3}{7}\right)$

At $(1,\,1)\!:\;\;\frac{dt}{dx} \:=\:\frac{2(2-1)}{1-7-6} \:=\:-\frac{1}{6}$

At $\left(1,\,-\tfrac{3}{7}\right)\!:\;\;\frac{dy}{dx} \:=\:\frac{2\left[2-(1)(-\frac{3}{7})\right]}{1 - 7(-\frac{3}{7}) - 6} \:=\:-\frac{17}{7}$

**derfleurer** · April 20th 2009, 11:15 AM

Whoop, missed that + 2 in the denominator.

Not to hijack the thread, but how'd you get 2xy' - 14yy' +2y' = 2(x - 7y - 6)y'?

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