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Derivatives of logarithmic functions Calculate the derivative of the following functions. In some cases, it is useful to use the properties of logarithms to simplify the functions before computing \(f^{\prime}(x)\). $$\begin{aligned} &4\\\ &f(x)=\ln \sqrt{10 x} \end{aligned}$$

Short Answer

Expert verified
Answer: The derivative of the given function is \(f'(x) = \frac{1}{2x}\).

Step by step solution

01

Simplify the function using logarithm properties

Given \(f(x) = \ln \sqrt{10 x}\). We can rewrite \(\sqrt{10x}\) as \((10x)^{\frac{1}{2}}\). Now, we can apply the logarithm property \(\ln a^b = b\ln a\). $$f(x) = \ln (10x)^{\frac{1}{2}} = \frac{1}{2}\ln(10x)$$ We'll now find the derivative of this simplified function.
02

Apply the chain rule

To find \(f'(x)\), we'll apply the chain rule, which states that if \(f(x) = g(h(x))\), then \(f'(x) = g'(h(x)) \cdot h'(x)\). Let \(g(u) = \frac{1}{2}\ln u\) and \(h(x) = 10x\). Then we have: $$f'(x) = g'(h(x)) \cdot h'(x)$$ Now we need to find \(g'(u)\) and \(h'(x)\).
03

Find \(g'(u)\)

To find \(g'(u)\), we know that \(g(u) = \frac{1}{2}\ln u\). The derivative of \(\ln u\) is \(\frac{1}{u}\), so the derivative of \(\frac{1}{2}\ln u\) is: $$g'(u) = \frac{1}{2}\cdot \frac{1}{u} = \frac{1}{2u}$$
04

Find \(h'(x)\)

To find \(h'(x)\), we know that \(h(x) = 10x\). The derivative of \(10x\) is: $$h'(x) = 10$$
05

Substitute \(g'(u)\) and \(h'(x)\) in the chain rule

Now, using the chain rule, we have: $$f'(x) = g'(h(x)) \cdot h'(x) = \frac{1}{2h(x)}\cdot 10 = \frac{1}{2(10x)}\cdot 10$$ Simplifying the expression, we get: $$f'(x) = \frac{10}{20x} = \frac{1}{2x}$$ Hence, the derivative of the given function is: $$f'(x) = \boxed{\frac{1}{2x}}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a crucial technique in calculus used for finding derivatives of composite functions. When a function is composed of an outer function and an inner function—like a Russian nesting doll—it's necessary to differentiate both layers. Here's a relatable way to think about it: imagine peeling an orange. To reach the fruit inside, you must first peel away the outer layer. Similarly, with the chain rule, you first differentiate the outer function, leaving the inner function untouched (think of it as the 'skin' of our orange), and then multiply it by the derivative of the inner function, or the 'fruit' inside.

As applied to our example, with an outer function \( g(u) = \frac{1}{2}\ln u \) and an inner function \( h(x) = 10x \), we first take the derivative of \( g(u) \) with respect to \( u \) and then multiply by the derivative of \( h(x) \) with respect to \( x \). Thus, the chain rule helps us tackle more complex functions by breaking them down into simpler parts, much like peeling an orange layer by layer.
Properties of Logarithms
Logarithms have special properties that can greatly simplify solving derivative problems when they're understood and applied correctly—think of them as tools in a toolbox, each meant for a specific job. For instance, one of these properties states that the logarithm of a power, like \( \ln(a^b) \), can be rewritten as the exponent multiplied by the logarithm of the base: \( b\ln a \). It's akin to rearranging building blocks to create a simpler structure before you start painting it—that is, before you differentiate it.

When we simplified \( f(x)=\ln \sqrt{10 x} \) to \( \frac{1}{2}\ln(10x) \), we used this power property to make our 'building' (function) easier to paint (differentiate). This approach often reduces the complexity of the function, making the differentiation process more straightforward.
Computing Derivatives
Computing derivatives is like taking a snapshot of the rate at which something changes. In our case, we wanted to capture the change of a logarithmic function. The derivative of a logarithm, specifically \( \ln u \) where \( u \) is a function of \( x \), is determined by the reciprocal of the inside function \( u \) times the derivative of \( u \). In simpler terms, you can think of it as calculating how 'sensitive' the logarithm is to changes in its input.

Once we had converted our function to \( \frac{1}{2}\ln(10x) \) using logarithmic properties, taking the derivative became a smooth sail: we just needed to find the 'reaction' of the logarithmic part to changes in \( 10x \) and then account for that initial factor of \(\frac{1}{2}\). This process is essential in many fields, from physics to economics, as it allows us to understand and predict the behavior of varying quantities.
Simplifying Logarithmic Expressions
Simplifying logarithmic expressions is like tidying up a room before a party—it makes it much easier to navigate and enjoy the space. In mathematics, this 'tidying up' can turn a daunting equation into one that's far more manageable. It usually involves applying properties of logarithms, such as the power rule, to rewrite the expression in a form that's easier to work with.

Our process of simplifying \( f(x)=\ln \sqrt{10 x} \) to \( \frac{1}{2}\ln(10x) \) before differentiating is a perfect example. We essentially arranged the 'furniture' in our 'mathematical room' to expedite our work, making it faster and more efficient to reach our goal—the derivative of the function. This simplification step is vital for students to master, as it can save time and reduce errors in more complex calculations.

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Most popular questions from this chapter

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x+y=\cos y$$

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