Question

How do you use implicit differentiation to find dy/dx given `2e^(3x)+e^(-2y)+7=0`?

Answer 1

`dy/dx=(6e^(3x))/(2e^(-2y))=(3e^(3x))/e^(-2y)`

Explanation:

`2e^(3x)+e^(-2y)+7=0`

`6e^(3x)-2e^(-2y)dy/dx=0`

`6e^(3x)=2e^(-2y)dy/dx`

`dy/dx=(6e^(3x))/(2e^(-2y))=(3e^(3x))/e^(-2y)`