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

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

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- Jun 11th 2010, 02:20 AMOnHoldDerative
How to find a derative by definition for such a function:

$\displaystyle y=xln^2(2x+1)$ - Jun 11th 2010, 02:26 AMcraig
- Jun 11th 2010, 02:35 AMOnHold
Yes, I have. I need to find derative in the point (3)

$\displaystyle ln^2(2h+2x+1)-(3/h)*ln^2(2x+1)$

$\displaystyle ln(2h+7)-ln(7)^(3/h)(ln(2h+7)+ln7^(3/h)$

There my knowledge finishes. - Jun 11th 2010, 04:39 AMProve It
You need to use the product rule and the chain rule...

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

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

So in your case:

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

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

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

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

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

Can you go from here?