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Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \sqrt{10 x}$$,

Short Answer

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Answer: The derivative of the function \(f(x) = \ln \sqrt{10x}\) is \(f'(x) = \frac{1}{2x}\).

Step by step solution

01

Simplify the function using logarithmic properties

First, we need to simplify the given function using the properties of the natural logarithm. Recall that \(\ln(\sqrt{x}) = \frac{1}{2}\ln(x)\) (which comes from the property \(\ln x^a = a \ln x\)), we can simplify the function as follows: $$f(x) = \ln \sqrt{10x} = \frac{1}{2} \ln (10x)$$
02

Compute the derivative using the chain rule

Now we will find the derivative \(f'(x)\) using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In our case, the outer function is \(\frac{1}{2} \ln u\) (where \(u = 10x\) is the inner function). The derivative of the outer function with respect to \(u\) is \(\frac{1}{2 u}\), and the derivative of the inner function with respect to \(x\) is \(10\). Applying the chain rule, we get: $$f'(x) = \frac{1}{2u} \cdot 10$$ Substitute \(u = 10x\) back into the equation: $$f'(x) = \frac{1}{2(10x)}\cdot 10$$
03

Simplify the derivative

Finally, we simplify the derivative: $$f'(x) = \frac{10}{20x} = \frac{1}{2x}$$ So the derivative of the given function is \(f'(x) = \frac{1}{2x}\).

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Key Concepts

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

Understanding Derivatives
A derivative measures how a function changes as its input changes. Think of it as the speed or rate of change of the function. For a small change in the input, how much does the output change?
For example, in the function \( f(x) = \ln \sqrt{10x} \), the derivative \( f'(x) \) tells us how quickly this logarithmic function is changing at any point \( x \).
  • The derivative provides valuable information about the behavior of the function around any given point.
  • It's especially useful in understanding the slope of the tangent line at specific points.
To calculate a derivative, we use different rules depending on the form of the function (like the chain rule, which we'll cover next). Knowing these rules allows us to handle various types of functions effectively. So, whenever you see \( f'(x) \), remember it's the rate at which \( f(x) \) is changing with respect to \( x \).
Chain Rule in Calculus
The chain rule is a powerful tool in calculus for finding the derivative of composite functions. A composite function is like nesting one function inside another, such as \( \ln(10x) \) embedded within our original function.

To apply the chain rule, follow these steps:
  • Identify the inner function (say, \( u = 10x \)) and the outer function (in this case, \( \frac{1}{2}\ln u \)).
  • Determine the derivative of the outer function with respect to the inner function. Here, it's \( \frac{1}{2u} \).
  • Find the derivative of the inner function with respect to \( x \), which is \( 10 \).
  • Multiply these derivatives together to get \( \frac{1}{2u} \cdot 10 \).
By substituting back \( u = 10x \), you get the complete derivative. The chain rule simplifies complex derivatives into a series of manageable steps, making it easier to handle nested functions like ours.
Natural Logarithm Properties
The natural logarithm, denoted as \( \ln \), is a special logarithm with the base \( e \), where \( e \approx 2.71828 \). It's commonly found in mathematics due to its unique properties, especially when dealing with growth processes or exponential decay.

In our example, simplifying \( \ln \sqrt{10x} \) utilizes the property \( \ln \sqrt{x} = \frac{1}{2}\ln x \). Here's how it works:
  • The square root of a number can be expressed as an exponent, \( x^{1/2} \).
  • The logarithmic identity \( \ln (x^a) = a \ln x \) allows us to bring the exponent in front as a multiplier.
This property transforms a potentially complex expression into a simpler one, facilitating easier differentiation or integration. The natural logarithm is not only fundamental in calculus but also appears in many real-world situations like banking, physics, and biology.

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Most popular questions from this chapter

A thin copper rod, 4 meters in length, is heated at its midpoint, and the ends are held at a constant temperature of \(0^{\circ} .\) When the temperature reaches equilibrium, the temperature profile is given by \(T(x)=40 x(4-x),\) where \(0 \leq x \leq 4\) is the position along the rod. The heat flux at a point on the rod equals \(-k T^{\prime}(x),\) where \(k>0\) is a constant. If the heat flux is positive at a point, heat moves in the positive \(x\) -direction at that point, and if the heat flux is negative, heat moves in the negative \(x\) -direction. a. With \(k=1,\) what is the heat flux at \(x=1 ?\) At \(x=3 ?\) b. For what values of \(x\) is the heat flux negative? Positive? c. Explain the statement that heat flows out of the rod at its ends.

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\log _{2} \frac{8}{\sqrt{x+1}}$$

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in \(\mathrm{km}\), the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}.\) a. Compute the pressure at the summit of Mt. Everest which has an elevation of roughly \(10 \mathrm{km}\). Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\). d. Does \(p^{\prime}(z)\) increase or decrease with \(z\) ? Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

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