Hi could someone please help me differentiate?

$\displaystyle

r(t) = ln(2t\sqrt(t+1))

$

I was thinking of using the product rule.

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- Feb 4th 2009, 07:20 AMstudentmeDerivative calculations
Hi could someone please help me differentiate?

$\displaystyle

r(t) = ln(2t\sqrt(t+1))

$

I was thinking of using the product rule. - Feb 4th 2009, 09:06 AMMoo
Hello,

It would come after the**chain**rule.

$\displaystyle [u(v(x))]'=v'(x)u'(v(x))$

Here, $\displaystyle u(y)=\ln(y)$ and $\displaystyle v(t)=2t \sqrt{t+1}$ and since you'll have to compute $\displaystyle v'(t)$, this is where you will have to use the product rule.

Otherwise, you can write it this way :

$\displaystyle r(t)=\ln(2t \sqrt{t+1})=\ln(2t)+\ln(\sqrt{t+1})=\ln(2)+\ln(t)+ \ln((t+1) {1/2})$ $\displaystyle =\ln(2)+\ln(t)+\frac 12 \ln(t+1)$

ln(2) is a constant, hence its derivative is 0.

ln(t) has a well-known derivative.

ln(t+1) has a simple derivative you can compute by using the chain rule.