Is it possible to differentiate this equation?
(2x^2)(y)-(7y^2)=(8x)-2y-11 to find y'(1)?
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
Hello, Hakhengkim!
Did you follow derfleurer's advice and finish the problem?
Differentiate implicitly . . .$\displaystyle 2x^2y-7y^2\:=\:8x-2y-11$. . Find $\displaystyle y'(1)$
. . $\displaystyle 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: .$\displaystyle 2(x-7y - 6)\,\frac{dy}{dx} \:=\:4(2-xy) \quad\Rightarrow\quad \frac{dy}{dx} \;=\;\frac{2(2-xy)}{x-7y-6}$
If $\displaystyle x = 1$, the equation becomes: .$\displaystyle 2y - 7y^2 \:=\:8 - 2y - 11 \quad\Rightarrow\quad 7y^2 - 4y - 3 \:=\:0$
. . $\displaystyle (y - 1)(7y+3) \:=\:0 \quad\Rightarrow\quad y \:=\:1,\:-\tfrac{3}{7}$
And we have two points: .$\displaystyle (1,\,1),\;\;\left(1,\,-\tfrac{3}{7}\right)$
At $\displaystyle (1,\,1)\!:\;\;\frac{dt}{dx} \:=\:\frac{2(2-1)}{1-7-6} \:=\:-\frac{1}{6}$
At $\displaystyle \left(1,\,-\tfrac{3}{7}\right)\!:\;\;\frac{dy}{dx} \:=\:\frac{2\left[2-(1)(-\frac{3}{7})\right]}{1 - 7(-\frac{3}{7}) - 6} \:=\:-\frac{17}{7}$