What is the slope of the tangent line of $(1 - x) (3 - 4 y^{2}) - ln y = C$ $(1-x)(3-4y^2)-lny = C$ , where C is an arbitrary constant, at $(1, 1)$ $(1,1)$ ?

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Answer 1 · 2016-08-09T18:22:19

$m = 1$ $m=1$

$f (x, y) = (1 - x) (3 - 4 y^{2}) - ln y - C$ $f(x,y)=(1-x)(3-4y^2)-lny-C$

$\frac{d y}{d x} = - \frac{f_{x}}{f_{y}}$ $dy/dx = - f_x/f_y$

$= - \frac{(- 1) (3 - 4 y^{2})}{(1 - x) (- 8 y) - \frac{1}{y}}$ $= -((-1)(3-4y^2))/((1-x)(-8y)-1/y)$

At $(1, 1)$ $(1,1)$ , we get

$\frac{d y}{d x} = - \frac{(- 1) (- 1)}{(0) (- 8) - 1} = 1$ $dy/dx = -((-1)(-1))/((0)(-8)-1) = 1$

What is the slope of the tangent line of (1-x)(3-4y^2)-lny = C (1−x)(3−4y2)−lny=C(1-x)(3-4y^2)-lny = C , where C is an arbitrary constant, at (1,1)(1,1)(1,1)?