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In this lesson, we will see how to use the quotient rule for derivatives to find the derivative of ln(x) / x. After we find out this derivative, we will see how to it might be used in a real world application.

Steps to Solve

Finding the derivative of the function h(x) = ln(x)/x all comes down to noticing that the function h(x) is a quotient of functions. That is, h(x) = f(x) / g(x), where f(x) = ln(x) and g(x) = x.

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Since we are dealing with a quotient of functions, we are going to make use of the quotient rule for derivatives – makes sense, right?The quotient rule for derivatives states that the derivative of f(x) / g(x) can be found using the formula (g(x)⋅f ‘ (x) – f(x);g ‘ (x)) / (g(x))2.

dervlnxx1


That’s a pretty big formula, but all we need to do is find the different parts involved in the formula, plug them in, and simplify. We can do that! As we said, in the function h(x) = ln(x) / x = f(x) / g(x), we have that f(x) = ln(x) and g(x) = x. All we need to know is the derivatives of both f and g, and we have all the parts we need to use the quotient rule. Thankfully, these derivatives are really well known!

  • The derivative of ln(x) is 1/x, so f ‘ (x) = 1/x.

  • The derivative of x is 1, so g ‘ (x) = 1.

Great! We have all our parts, now let’s plug them into the quotient rule and find the derivative of ln(x) / x.

dervlnxx2


We see that the derivative of h(x) = ln(x) / x is (1 – ln(x)) / x2. That quotient rule really makes this problem fairly easy to solve!

Solution

The derivative of ln(x) / x is (1 – ln(x)) / x2.

Application

Now that we know how to find the derivative of ln(x) / x, let’s talk about how this new-found knowledge might be useful. That is, let’s talk about sugar! Wait, what? You might be wondering how this problem relates to sugar.

Well, let’s find out!You may know that when we eat a sugary snack, our blood sugar spikes and then slowly goes back down. Well, take a look at the graph of h(x) = ln(x) / x.

dervlnxx3


See how the graph increases quickly, or spikes, and then slowly goes back down towards zero? Well, this is exactly what we just said our blood sugar does when we eat a sugary snack!Suppose you just ate something deliciously sweet, and the spike in your blood sugar is modeled by our function h(x) = ln(x) / x, where h(x) is the amount of change in your blood sugar (in millimoles per liter) after you eat your snack, and x is the number of minutes that have passed since you ate the snack.The derivative of this function, which we just found to be h ‘ (x) = (1 – ln(x)) / x2, gives us the rate at which this function is increasing or decreasing. In other words, the derivative can give us the rate at which your blood sugar is rising or falling after x minutes.

For instance, if you wanted to know the rate at which your blood sugar was changing one minute after you ate your snack, you simply plug x = 1 into the derivative formula.

dervlnxx4


We see that 1 minute after you had your snack, your blood sugar is rising quite quickly at 1 mmo/L per minute.

Now, suppose you want to know the same thing, but at 5 minutes after you ate your snack. No problem! Once again, we simply plug x = 5 into our derivative formula to find this rate.

dervlnxx5


We see that 5 minutes after you ate your snack, the rate of change of your blood sugar is approximately -0.

024 mmo/L per minute. The fact that it’s negative tells us that you’ve already experienced your spike, and your blood sugar is slowly going down now.Pretty neat that we can get all this information just by knowing the derivative of the function ln(x) / x, huh? Now that we know how to find this derivative, we can always find the rate of change of any phenomena that can be modeled by the function h(x) = ln(x) / x. I don’t know about you, but I’d say that’s some pretty useful information!

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