Question

Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{dy}{dx}$

$(3x+1)(y^2-y+6)=0$

Short answer

$\frac{dy}{dx}=\frac{(-3)(y^2-y+6)}{(3x+1)(2y-1)}$

Step-by-step solution

Step 1. Given Information:

Given equation:$(3x+1)(y^2-y+6)=0$

We want to find $\frac{dy}{dx}$ defines y as an implicit function of x by use implicit differentiation.

Step 2. Solution: 

Differentiate both sides w.r.t. x

$\frac{d}{dx}(3x+1)(y^2-y+6)=\frac{d}{dx}0$

Using product and chain rule we get

$$\begin{gathered} (3x+1)\frac{d}{dx}(y^2-y+6)+(y^2-y+6)\frac{d}{dx}(3x+1)=\frac{d}{dx}0 \\ (3x+1)(2y·\frac{dy}{dx}-\frac{dy}{dx}+0)+(y^2-y+6)\left(3\right)=0 \\ (3x+1)(2y-1)\frac{dy}{dx}=(-3)(y^2-y+6) \\ \frac{dy}{dx}=\frac{(-3)(y^2-y+6)}{(3x+1)(2y-1)} \end{gathered}$$