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

Feb 3, 2016

The slope is $1$

#### Explanation:

This looks too tedious as written, so I'll rewrite.

Given that $\left(1 , 1\right)$ lies on the curve, $C$ is not arbitrary, but can be found:

$\frac{1 - \left(1\right)}{3 - 4 {\left(1\right)}^{2}} - \left(1\right) = C$, so $C = - 1$.

Therefore, if we replace $C$ by $- 1$ and multiply through by $3 - 4 {y}^{2}$, we get

$1 - x - 3 y + 4 {y}^{3} = 4 {y}^{2} - 3$.

Differentiating implicitly with respect to $x$ gets us

$- 1 - 3 \frac{\mathrm{dy}}{\mathrm{dx}} + 12 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 8 y \frac{\mathrm{dy}}{\mathrm{dx}}$.

Now, find $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(1 , 1\right)$

$- 1 + \left(- 3 + 12 - 8\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$