# In is (1/2)x(-1/2), or 1/(2;x). These facts will

In this video, we’ll learn how to find the derivative of ln(sqrt(x)) and review the chain rule for derivatives.

After we’ve found the derivative, we’ll see how it can be applied to an everyday situation.

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## Steps to Solve

If we want to find the derivative of ln(;x), we will have to make use of the chain rule for derivatives. The chain rule for derivatives is a rule that we use to find the derivative of functions of the form f(g(x)). Notice that if we let f(x) = ln(x) and g(x) = ;x, then f(g(x)) = ln(;x), which tells us that we can use the chain rule to find the derivative of ln(;x).

Let’s think about what else we’ll need to know to find this derivative. The chain rule states that if h(x) = f(g(x)), then h ‘ (x) = f ‘ (g(x)) ; g ‘ (x)In our example, f(x) = ln(x) and g(x) = √x. To use the chain rule, we’re going to have to find f ‘ (g(x)), and g ‘ (x). Both of these derivatives are well-known.

• The derivative of ln(x) is 1 / x.

• The derivative of ;x is (1/2)x(-1/2), or 1/(2;x).

These facts will be helpful in our quest for the derivative. Since the derivative of ln(x) is 1/x, we have that f ‘ (x) = 1/x, so f ‘ (g(x)) = 1/(;x). Also, we know that g ‘ (x) = 1 / (2√x). So all we have to do is plug these into the chain rule formula and simplify to get our derivative.

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## Application

Finding the derivative of ln(√x) wasn’t so hard. Now, let’s consider a scenario where being able to do so might come in handy. Suppose a girl named Mandi has gone through a growth spurt over the past few years. We can model her growth over these 36 months by using the function h(x) = ln(√x) + 1, where h(x) is the number of inches that she’s grown during this growth spurt, and x is the number of months since the growth spurt started.Mandi notices that when her growth spurt first began, she grew very quickly, but it seems to have slowed, though she’s still growing. We can illustrate this trend by graphing the function. This gets Mandi to thinking about just how quickly she was growing at different times over the past 36 months. We can use the derivative of ln(√x) to answer her question.As you may know, the derivative of a function represents the rate at which y is changing with respect to x at any given value of x, so we just need to find the derivative of our function h(x) = ln(√x) + 1, and we’ll have a formula to find Mandi’s growth rate at any time since the start of her spurt.To do this, we can make use of the fact that the derivative of a sum is the sum of the derivatives, and that the derivative of a constant is 0. Thus, the derivative of h(x) is the sum of the derivative of ln(√x) and the derivative of 1.

We know the derivative of ln(√x) is 1/(2x). Putting all this information together, we get the following:h ‘ (x) = 1/(2x) + 0 = 1/(2x)We now have a formula that will give us the rate at which Mandi was growing x months after the growth spurt started.For example, if she wants to know what her growth rate was four months in, she simply needs to plug x = 4 into the derivative formula.1 / (2x) = 1 / (2;4) = 1/8Four months after Mandi’s growth spurt started, she was growing at a rate of 1/8 of an inch each month. Similarly, if she wants to know what her growth rate was 30 months in, she would plug x = 30 into the derivative formula.1/ (2x) = 1 / (2⋅30) = 1/60Here Mandi was growing at a rate of 1/60 of an inch per month at 30 months in.

So she was right about the fact that she grew more quickly at the beginning of her growth spurt.Growth is an ordinary human process that we all experience and we can use the derivative of ln(√x) to analyze it.

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