# Logarithmic function

• Dec 5th 2009, 06:08 PM
Archduke01
Logarithmic 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 PM
Quote:

Originally Posted by Archduke01
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?

You have to use implicit differentiation. This should be easy, I think it is the $ln(y^2)$ term that is confusing you.

One of the properties of logarithms allows you to write:

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

Use the chain rule to differentiation this:

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

That should help.
• Dec 5th 2009, 06:16 PM
Archduke01
Quote:

You have to use implicit differentiation. This should be easy, I think it is the $ln(y^2)$ term that is confusing you.

One of the properties of logarithms allows you to write:

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

Use the chain rule to differentiation this:

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

That should help.

What does the 2y and the 2x become?
• Dec 5th 2009, 07:04 PM
Archduke01
Quote:

Originally Posted by Archduke01
What does the 2y and the 2x become?

Does the 2y become 2yy'? What does the 2x become?