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

Aug 9, 2016

$m = 1$

#### Explanation:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

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

At $\left(1 , 1\right)$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\left(- 1\right) \left(- 1\right)}{\left(0\right) \left(- 8\right) - 1} = 1$