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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 dydx

(3x+1)(y2y+6)=0

Short Answer

Expert verified

dydx=(-3)(y2-y+6)(3x+1)(2y-1)

Step by step solution

01

Step 1. Given Information:

Given equation:(3x+1)(y2y+6)=0

We want to find dydx defines y as an implicit function of x by use implicit differentiation.

02

Step 2. Solution: 

Differentiate both sides w.r.t. x

ddx(3x+1)(y2y+6)=ddx0

Using product and chain rule we get

(3x+1)ddx(y2y+6)+(y2y+6)ddx(3x+1)=ddx0(3x+1)(2y·dydx-dydx+0)+(y2-y+6)(3)=0(3x+1)(2y-1)dydx=(-3)(y2-y+6)dydx=(-3)(y2-y+6)(3x+1)(2y-1)

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