/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Find \(d^{2} y / d x^{2}\) $$y... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 51

Find \(d^{2} y / d x^{2}\) $$y=x \cos (5 x)-\sin ^{2} x$$

Short Answer

Expert verified
The second derivative is \(-10\sin(5x) - 25x\cos(5x) + 2\cos(2x)\)."

Step by step solution

01

Identify Components for Differentiation

The function given is \(y = x \cos(5x) - \sin^2(x)\). We need to take the second derivative of this function with respect to \(x\), which requires finding the first derivative \(dy/dx\) first. Notice that the expression has two main components: \(x \cos(5x)\) and \(-\sin^2(x)\).
02

Differentiate the first component

Start by differentiating \(x \cos(5x)\). Use the product rule for differentiation, which states \(d(uv)/dx = u'v + uv'\). Let \(u = x\) and \(v = \cos(5x)\), then \(u' = 1\) and \(v' = -5\sin(5x)\). So, \(d(x \cos(5x))/dx = 1 \cdot \cos(5x) + x(-5\sin(5x)) = \cos(5x) - 5x\sin(5x)\).
03

Differentiate the second component

Now differentiate \(-\sin^2(x)\). Use the chain rule: if \(f(x) = -\sin^2(x)\), then let \(u = \sin(x)\) so \(f = -u^2\). The derivative is \(df/dx = d(-u^2)/du \, \cdot \, du/dx = -2u \cdot \cos(x) = -2\sin(x)\cos(x)\).
04

Combine first derivatives

Combine the results from Steps 2 and 3 to find the first derivative: \(dy/dx = \cos(5x) - 5x\sin(5x) - 2\sin(x)\cos(x)\).
05

Differentiate again for the second derivative

Differentiate the expression from Step 4 to find \(d^2y/dx^2\). Use the sum rule and product rule where necessary. Differentiate each term separately: 1. \(d/dx[\cos(5x)] = -5\sin(5x)\).2. \(d/dx[-5x\sin(5x)] = -5[1 \cdot \sin(5x) + x \cdot 5\cos(5x)] = -5\sin(5x) - 25x\cos(5x)\).3. \(d/dx[-2\sin(x)\cos(x)] = -2[-\sin^2(x) + \cos^2(x)] = 2(\cos^2(x) - \sin^2(x)) = 2\cos(2x)\) using the double-angle identity.
06

Combine second derivative components

Combine the results from Step 5 to get the second derivative: \(\frac{d^2y}{dx^2} = -5\sin(5x) - 5\sin(5x) - 25x\cos(5x) + 2\cos(2x) = -10\sin(5x) - 25x\cos(5x) + 2\cos(2x)\).

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Key Concepts

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

Product Rule
The product rule is a cornerstone of calculus used when differentiating products of two functions. When given a function of the form \(y = u(x)v(x)\), the derivative \(dy/dx\) is found using the formula
  • \(\frac{d}{dx}[uv] = u'v + uv'\).
In the context of the original exercise, the term \(x\cos(5x)\) required us to apply the product rule. Here, we defined \(u = x\) and \(v = \cos(5x)\). Consequently, the derivatives were \(u' = 1\) and \(v' = -5\sin(5x)\).
This led to:
  • \(\frac{d}{dx}[x\cos(5x)] = 1 \cdot \cos(5x) + x (-5\sin(5x)) = \cos(5x) - 5x\sin(5x)\).
This step shows how combining the derivatives of \(u\) and \(v\) correctly contributes to finding \(dy/dx\) accurately.
Chain Rule
The chain rule is instrumental when differentiating compositions of functions. It explains how to find the derivative of a function \(f(g(x))\) as
  • \(f'(g(x)) \cdot g'(x)\).
In our exercise, we needed the chain rule for \(-\sin^2(x)\). Letting \(u = \sin(x)\) transformed the expression into \(-u^2\). This required
  • \(\frac{d}{dx}[-\sin^2(x)] = -2u \cdot \frac{du}{dx} = -2\sin(x) \cdot \cos(x)\).
The chain rule simplifies differentiations where basic algebraic manipulations and direct differentiation rules cannot be applied straightforwardly. This helps achieve the correct first derivative component for \(-\sin^2(x)\).
Trigonometric Functions
Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) are common in calculus, and understanding their derivatives is vital. For instance, the derivatives are:
  • \(\frac{d}{dx}[\sin(x)] = \cos(x)\)
  • \(\frac{d}{dx}[\cos(x)] = -\sin(x)\)
In our solution, these properties helped differentiate terms like \(\cos(5x)\) and those involving products like \(-\sin^2(x)\). The term \(\cos(2x)\) derived during the second differentiation results from trigonometric identities like the double-angle formula:
  • \(\cos(2x) = \cos^2(x) - \sin^2(x)\).
Such identities are useful in simplifying and accurately transforming expressions that's needed to complete tasks like finding \(d^2y/dx^2\) in a manageable way.

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Most popular questions from this chapter

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{3}+\frac{1}{16}, & x<\frac{1}{2} \\ \frac{2}{4} x^{2}, & x \geq \frac{1}{2} \end{array}\right. $$ Determine whether \(f\) is differentiable at \(x=\frac{1}{2} .\) If so, find the value of the derivative there.

Use a CAS to find dy/dx. $$y=\tan ^{4}\left(2+\frac{(7-x) \sqrt{3 x^{2}+5}}{x^{3}+\sin x}\right)$$

Given the following table of values, find the indicated derivatives in parts (a) and (b). $$\begin{array}{|c|c|c|} \hline x & f(x) & f^{\prime}(x) \\ \hline 2 & 1 & 7 \\\ \hline 8 & 5 & -3 \\ \hline \end{array}$$ (a) \(g^{\prime}(2),\) where \(g(x)=[f(x)]^{3}\) (b) \(h^{\prime}(2),\) where \(h(x)=f\left(x^{3}\right)\)

Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=x^{2} \sqrt{5-x^{2}}, x=1$$

Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=\sin \left(1+x^{3}\right), x=-3$$
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