Deriving 4x^3 + ln y^2 + 2y = 2x

I think the first step is; 12x^2 + 1 / y^2 + and here is where I'm stuck. What is the derivative of the 2y and 2x?

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- Dec 5th 2009, 06:08 PMArchduke01Logarithmic function
Deriving 4x^3 + ln y^2 + 2y = 2x

I think the first step is; 12x^2 + 1 / y^2 + and here is where I'm stuck. What is the derivative of the 2y and 2x? - Dec 5th 2009, 06:13 PMadkinsjr
You have to use implicit differentiation. This should be easy, I think it is the $\displaystyle ln(y^2)$ term that is confusing you.

One of the properties of logarithms allows you to write:

$\displaystyle ln(y^2)=2ln(y)$

Use the chain rule to differentiation this:

$\displaystyle 2\frac{d}{dx}ln(y)=\frac{2}{y}\frac{dy}{dx}$

That should help. - Dec 5th 2009, 06:16 PMArchduke01
- Dec 5th 2009, 07:04 PMArchduke01