Derative

**OnHold** · June 11th 2010, 02:20 AM

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

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

**craig** · June 11th 2010, 02:26 AM

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?

**OnHold** · June 11th 2010, 02:35 AM

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.

**Prove It** · June 11th 2010, 04:39 AM

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

So in your case:

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

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?

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