1. ## Derative

How to find a derative by definition for such a function:

$y=xln^2(2x+1)$

2. Originally Posted by OnHold
How to find a derative by definition for such a function:

$y=xln^2(2x+1)$
Have you made any attempt at this, using the product and the chain rule?

3. Yes, I have. I need to find derative in the point (3)

$ln^2(2h+2x+1)-(3/h)*ln^2(2x+1)$
$ln(2h+7)-ln(7)^(3/h)(ln(2h+7)+ln7^(3/h)$
There my knowledge finishes.

4. Originally Posted by OnHold
How to find a derative by definition for such a function:

$y=xln^2(2x+1)$
You need to use the product rule and the chain rule...

Chain rule: $\frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$.

Product rule: $\frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}$.

$y = x[\ln{(2x + 1)}]^2$.

Can you see that this is a product of functions? So you need to use the product rule with $u = x$ and $v = [\ln{(2x + 1)}]^2$.

So $\frac{dy}{dx} = x\,\frac{d}{dx}\{[\ln{(2x + 1)}]^2\} + [\ln{(2x + 1)}]^2\,\frac{d}{dx}(x)$

$= x\,\frac{d}{dx}\{[\ln{(2x + 1)}]^2\} + [\ln{(2x + 1)}]^2$.

Now to find $\frac{d}{dx}\{[\ln{(2x + 1)}]^2\}$ you need to use the chain rule twice...

Can you go from here?