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Chapter 3: Problem 2

\(2-22\) Differentiate the function. \(f(x)=\ln \left(x^{2}+10\right)\)

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{2x}{x^2 + 10} \).

Step by step solution

01

Identify the Differentiation Rule

The function is \( f(x) = \ln(x^2 + 10) \). To differentiate this, we'll use the chain rule combined with the derivative of the natural logarithm, \( d/dx \ln(u) = 1/u \, (du/dx) \).
02

Differentiate the Outer Function

The outer function is \( \ln(u) \) where \( u = x^2 + 10 \). The derivative of \( \ln(u) \) with respect to \( u \) is \( 1/u \). So, we have \( 1/(x^2 + 10) \).
03

Find the Derivative of the Inner Function

Now, we need to differentiate the inner function \( u = x^2 + 10 \) with respect to \( x \). The derivative \( du/dx \) is \( 2x \).
04

Apply the Chain Rule

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function. This gives us \( 1/(x^2 + 10) \times 2x \), or simply \( \frac{2x}{x^2 + 10} \).
05

Write the Final Derivative

The derivative of the function \( f(x) = \ln(x^2 + 10) \) is \( f'(x) = \frac{2x}{x^2 + 10} \).

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Key Concepts

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

Chain Rule
When differentiating compositions of functions, like the one in the exercise, the chain rule becomes essential. The chain rule states that to differentiate a composite function \( f(g(x)) \), you need to differentiate the outer function \( f \) with respect to the inner function \( g(x) \), and then multiply this result by the derivative of the inner function \( g(x) \). By applying this rule, you navigate through each layer of the function separately.
  • Identify the inner function \( u = g(x) \).
  • Differentiate the outer function \( f(u) \) to get \( f'(u) \).
  • Differentiate the inner function to find \( g'(x) \).
  • Combine them as \( f'(u) \cdot g'(x) \).
This method is remarkably effective for breaking down complex problems. In our exercise, the chain rule allowed us to handle the natural logarithm of a quadratic expression by treating it as a layered structure.
Natural Logarithm Differentiation
Differentiating natural logarithms requires understanding the properties of the natural log function, \( \ln(x) \). The key rule to remember is that the derivative of \( \ln(u) \) is \( 1/u \) times the derivative of \( u \) with respect to \( x \), expressed as \( \frac{d}{dx}\ln(u) = \frac{1}{u} \frac{du}{dx} \).
  • Logarithms simplify multiplication and division, making them easier to differentiate.
  • Start by identifying the expression inside the logarithm \( u \).
  • Calculate \( \frac{1}{u} \).
  • Multiply by \( \frac{du}{dx} \).
In our exercise, using this rule on \( \ln(x^2 + 10) \), we found \( u = x^2 + 10 \), leading to the expression \( 1/(x^2 + 10) \) as part of the final derivative.
Derivative of a Function
Understanding derivatives is fundamental in calculus. A derivative represents how a function changes as its input changes; it鈥檚 essentially the slope of the function at any point. Calculating derivatives involves using a variety of rules and techniques, depending on the function type.
  • Derivatives are key in finding rates of change in various contexts, like velocity and acceleration in physics.
  • A basic derivative result is that \( \frac{d}{dx}x^n = nx^{n-1} \).
  • Complex functions require combining rules like the product, quotient, and chain rules.
  • Differentiation helps to optimize functions and find local maxima or minima.
In the given task, we evaluated the rate of change for \( f(x) = \ln(x^2 + 10) \) using the chain rule and properties of logarithms, resulting in a succinct expression for \( f'(x) \). Understanding derivatives helps translate real-world problems into solvable mathematical expressions.

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Most popular questions from this chapter

Find \(\frac{d^{9}}{d x^{9}}\left(x^{8} \ln x\right)\)

The edge of a cube was found to be 30 \(\mathrm{cm}\) with a possible error in measurement of 0.1 \(\mathrm{cm}\) . Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area of the cube.

$$ \begin{array}{l}{\text { The radius of a circular disk is given as } 24 \mathrm{cm} \text { with a maxi- }} \\ {\text { mum error in measurement of } 0.2 \mathrm{cm} .} \\ {\text { (a) Use differentials to estimate the maximum error in the }} \\ {\text { calculated area of the disk. }} \\ {\text { (b) What is the relative error? What is the percentage error? }}\end{array} $$

Use the definition of derivative to prove that \(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1\)

The frequency of vibrations of a vibrating violin string is given by $$\mathrm{f}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\rho}}$$ where \(L\) is the length of the string, T is its tension, and \(\rho\) is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3 \(\mathrm{d}\) ed. (Pacific Grove, CA: Brooks/Cole, \(2002 ) . ]\) (a) Find the rate of change of the frequency with respect to (i) the length (when T and \(\rho\) are constant), (ii) the tension (when L and \(\rho\) are constant), and (iii) the linear density (when L and T are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency f. (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg. (iii) when the linear density is increased by switching to another string.
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