LN word problem measurement of a child's ability to learn is given by the function L(t)=(ln(t+1))/(t+1) where t is the child's age in years, for 0<=t<=5 At what age does a child have the greatest learning ability? At what age is a child's learning ability increasing most rapidly? Honestly, I'm not even sure where to start, should I take the 2nd derivative?

1. At what age does a child have the greatest learning abilit [sic.] ?
The child's ability to learn is measured by the function, so the first question asks when that ability is maximal, or when the function attains its maximum value. That is done by taking the first derivative and performing the first derivative test, and also testing at the endpoints, like so:
$$L(t)=\frac{ln(t+1)}{t+1}$$
$$L^{\prime }(t)=\frac{1-<!--\; -\; -->ln(t+1)}{(t+1{)}^2}$$
Solving for the critical numbers only requires checking the numerator and denominator separately, since if either is 0 there will be an undefined or zero value for the function, either way a critical number.
$1-<!--\; -\; -->ln(t+1)=0$ attains a zero where the natural log 1 equals 1, so at e - 1 or about 1.71828
$(t+1{)}^2=0$ has no solutions on the domain $0\le <!--\; \le \; -->t\le <!--\; \le \; -->5$
So which is the maximum? We simply check the values and receive x = 1.71828 as our maximum.
L(0) = 0 L(1.71828) = 0.36788 L(5) = 0.29863
Therefore, the child has the greatest learning ability at about 1.71828 years of age.
2. At what age is a child's learning ability increasing most rapidly?
The rate of change of the increase of the function is a rate of change of a rate of change: simply find where the first derivative attains its maximum, by using the first derivative test on the first derivative. That means you do have to find the second derivative as you assumed, and do the same test I demonstrated on that. Although it might be a bit longer, the concept isn't any trickier so I leave it to you. Be sure to check the endpoints!