# How do you find the slope of the tangent line to the curve 4xy^3+3xy=7 at the point (1,1) by using imlicit differentiation?

Aug 20, 2015

I found: Slope$= - \frac{7}{15}$

#### Explanation:

First you need to derive implicitly (rememberin that $y$ will be a function of $x$) and get (using the Product Rule on the two terms on the left):
$4 {y}^{3} + 12 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 y + 3 x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 {y}^{3} - 3 y}{12 x {y}^{2} + 3 x} =$Slope
Now use the coordinates of your point:
$x = 1$
$y = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 - 3}{12 + 3} = - \frac{7}{15}$