Logarithmic differentiation

Hedgecutter

Here's a question I'm trying at the moment. The blue writhing is my work and red is the answer from the book.

I don't understand where they get 6/(2x+1), I got 3/(2x+1).

Pherekydes

Don't forget the derivative of the bracket.

Hedgecutter

Pherekydes wrote: »

Don't forget the derivative of the bracket.

Derivative of 2x+1 is 2, do I multiply that by 3 = 6 ?

Pherekydes

Yes.

Hedgecutter

Pherekydes wrote: »

Yes.

So do you always work out the derivative of what's in the brackets (under the line) and multiply by what's above the line?

Moody_mona

The chain rule is being used here. The differentiation of log(f(x)) = (1/f(x)) times f'(x)

Yakuza

As Mona above says, it's the Chain rule : dy/dx = dy/du * du/dx.
y = ln(2x+1), set u = 2x+1, du/dx = 2
y = ln(u) so dy/du = 1/u
dy/dx = 1/u * 2 or 2/(2x+1)

In the original question the function is 3 * ln(2x+1), so the answer is 3 * (2 / (2x +1) (by the product rule - d (uv)/dx = v du/dx + u dv/dx but one of the terms is a constant so its derivative is zero)