At what rate is this function increasing?

IntegrateMe
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When x = 16,the rate at which \sqrt x is increasing is \frac {1}{k} times the rate at which x is increasing. What is the value of k?
 
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I thought it would be 4 but the answer is 8.
 
Its the ratio of the derivatives evaluated at x=16,

(2 sqrt(x))^-1
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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