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Chapter 3: Problem 23

\(23-26 \text { Find } y^{\prime} \text { and } y^{\prime \prime}\) \(y=x^{2} \ln (2 x)\)

Short Answer

Expert verified
\(y' = 2x \ln(2x) + x\) and \(y'' = 2 \ln(2x) + 3\).

Step by step solution

01

Differentiate to find \(y'\)

To find the first derivative \(y'\) of the function \(y = x^2 \ln(2x)\), we use the product rule. The product rule states that if you have a function in the form \(u(x) \cdot v(x)\), then \((uv)' = u'v + uv'\).Here, \(u = x^2\) and \(v = \ln(2x)\).First, compute \(u'\):\[ u' = \frac{d}{dx}(x^2) = 2x \]Next, compute \(v'\) using the chain rule:\[ v' = \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \]Now, apply the product rule:\[ y' = u'v + uv' = (2x) \cdot \ln(2x) + x^2 \cdot \frac{1}{x} \]Simplify the expression:\[ y' = 2x \ln(2x) + x \]
02

Differentiate again to find \(y''\)

To find the second derivative \(y''\), differentiate \(y' = 2x \ln(2x) + x\) again. Apply the product rule to \(2x \ln(2x)\):In \(2x \ln(2x)\), let \(u = 2x\) and \(v = \ln(2x)\), giving us the product rule:For \(u = 2x\), \(u' = 2\).For \(v = \ln(2x)\), \(v' = \frac{1}{x}\) as before.the product is:\[ (2x \ln(2x))' = u'v + uv' = 2 \cdot \ln(2x) + 2x \cdot \frac{1}{x} = 2 \ln(2x) + 2 \]So, \(y'' = (2x \ln(2x))' + x'\), where \(x' = 1\).Combine the parts and simplify:\[ y'' = 2 \ln(2x) + 2 + 1 = 2 \ln(2x) + 3 \]
03

Review the results

We have calculated the first derivative as \(y' = 2x \ln(2x) + x\) and the second derivative as \(y'' = 2 \ln(2x) + 3\). These results are the final answers for the problem.

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Key Concepts

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

Product Rule
When dealing with calculus differentiation, the product rule is essential when the function consists of multiples of two differentiable functions. In our original problem, we have:
  • Function 1: \(x^2\)
  • Function 2: \(\ln(2x)\)
The product rule formula is:\[(uv)' = u'v + uv'\]This helps us determine the derivative of the product of two functions. Here's how it鈥檚 applied:- Derivatives of each part are found separately.- Finally, the formula above combines them.
This method is powerful because it helps simplify the breakdown of complex functions. You can focus on smaller pieces and combine them afterward for the complete derivative.
Chain Rule
The chain rule is a vital tool in calculus, especially when functions are composed together. In our problem, it was used to differentiate \(\ln(2x)\), which involves an internal function within another one.
The chain rule formula is:\[(\ln(2x))' = \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot \frac{d}{dx}(2x)\]Instead of directly differentiating, use the outside function, then multiply by the derivative of the inside. Here:
  • Outer function: \(\ln(u)\)
  • Inner function: \(u = 2x\)
  • The derivative of \(u = 2\)
The result from applying this would be \(\frac{1}{x}\), combining the derivatives of inner and outer functions.
Second Derivative
Calculating the second derivative involves differentiating the first derivative you already found. This process gives insights into the curvature or concavity of the function. For the given problem, the first derivative was:\[y' = 2x \ln(2x) + x\]To find the second derivative:
  • Differentiate each term again.
  • Apply product rule where necessary, such as with \((2x\ln(2x))'\).
For example, differentiating \(2x \ln(2x)\) involves:\[2 \ln(2x) + 2\]Adding these results with the derivative of the last term gives:\[y'' = 2 \ln(2x) + 2 + 1 = 2 \ln(2x) + 3\]The second derivative simplifies understanding how the function behaves beyond just the slope, showing how it bends.

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Most popular questions from this chapter

A water trough is 10 \(\mathrm{m}\) long and a cross-section has the shape of an isosceles trapezoid that is 30 \(\mathrm{cm}\) wide at the bottom, 80 \(\mathrm{cm}\) wide at the top, and has height 50 \(\mathrm{cm} .\) If the trough is being filled with water at the rate of 0.2 \(\mathrm{m}^{3} / \mathrm{min}\) , how fast is the water level rising when the water is 30 \(\mathrm{cm}\) deep?

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 \(\mathrm{m}\) higher than the bow of the boat. If the rope is pulled in at a rate of \(1 \mathrm{m} / \mathrm{s},\) how fast is the boat approaching the dock when it is 8 \(\mathrm{m}\) from tho dock?

Use the Chain Rule to show that if \(\theta\) is measured in degrees, then $$\frac { d } { d \theta } ( \sin \theta ) = \frac { \pi } { 180 } \cos \theta$$ This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)

\(37-48\) Use logarithmic differentiation to find the derivative of the function. $$y=(\tan x)^{1 / x}$$

Find the derivative. Simplify where possible. $$f(t)=\operatorname{csch} t(1-\ln \csc h t)$$
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