Derivative calculations

Hello,

Quote:

Originally Posted by studentme

Hi could someone please help me differentiate?

$<br /> r(t) = ln(2t\sqrt(t+1))<br />$

I was thinking of using the product rule.

It would come after the chain rule.

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

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

Otherwise, you can write it this way :
$r(t)=\ln(2t \sqrt{t+1})=\ln(2t)+\ln(\sqrt{t+1})=\ln(2)+\ln(t)+ \ln((t+1) {1/2})$ $=\ln(2)+\ln(t)+\frac 12 \ln(t+1)$
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.