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Chapter 3: Problem 4

In Exercises \(1-12,\) find \(d y / d x\) $$ y=x^{2} \cot x-\frac{1}{x^{2}} $$

Short Answer

Expert verified
\( \frac{dy}{dx} = 2x \cot x - x^2 \csc^2 x + \frac{2}{x^3} \).

Step by step solution

01

Differentiate Each Term

To find the derivative \( \frac{dy}{dx} \) of the function \( y = x^2 \cot x - \frac{1}{x^2} \), we first differentiate each term separately. We use the product rule for the first term \( x^2 \cot x \) and the power rule for the second term \(-\frac{1}{x^2} \).
02

Apply the Product Rule for \( x^2 \cot x \)

The product rule states that if \( u = x^2 \) and \( v = \cot x \), then \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x^2 \) and \( v = \cot x \). So, \( u' = 2x \) and \( v' = -\csc^2 x \). Thus, \( \frac{d}{dx} (x^2 \cot x) = 2x \cot x + x^2(-\csc^2 x) \).
03

Differentiate \(-\frac{1}{x^2}\) Using the Power Rule

The power rule is used for terms like \(-\frac{1}{x^2} = -x^{-2} \). Thus, the derivative of \(-x^{-2} \) is \( -(-2)x^{-3} = \frac{2}{x^3} \).
04

Combine the Derivatives

Combine the derivatives obtained from Steps 2 and 3. The derivative of the function becomes: \[ \frac{dy}{dx} = 2x \cot x - x^2 \csc^2 x + \frac{2}{x^3} \].
05

Simplify the Expression

If possible, simplify the expression for \( \frac{dy}{dx} \) to make it easier to interpret. In this case, the expression is already simplified: \[ \frac{dy}{dx} = 2x \cot x - x^2 \csc^2 x + \frac{2}{x^3} \].

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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 functions that are products of two smaller functions, like \( x^2 \cot x \), we rely on the product rule to differentiate them. The product rule is a straightforward method to find the derivative of a product of two functions. It says: \(\frac{d}{dx}(uv) = u'v + uv'\).
Here, \( u = x^2 \) and \( v = \cot x \). We need the derivatives of these functions independently:
  • \(u' = 2x\)
  • \(v' = -\csc^2 x\)
We apply the product rule by substituting these into the formula, yielding:\[ \frac{d}{dx} (x^2 \cot x) = 2x \cot x + x^2(-\csc^2 x) \].
Using the product rule helps break down complex differentiations into manageable steps, simplifying the process of finding derivatives for combined functions.
Power Rule
The power rule is one of the most basic yet powerful tools in calculus. It allows us to find derivatives of any function of the form \( x^n \), where \( n \) is a real number. The rule states:
If \( y = x^n \), then \( \frac{d}{dx}(x^n) = nx^{n-1} \).
In our problem, we have the function part \(-\frac{1}{x^2} = -x^{-2} \), which fits perfectly to apply the power rule.
  • The derivative of \(-x^{-2}\) is calculated as: \(-(-2)x^{-3} = \frac{2}{x^3} \).
The power rule simplifies the differentiation of polynomial terms, making it a fundamental tool in calculus. It helps us transform power expressions quickly, reducing errors in calculations.
Trigonometric Derivatives
Trigonometric functions have unique derivatives that are essential in calculus, particularly when differentiating expressions involving them.
  • Derivative of \( \cot x \): The derivative of \( \cot x \) is \(-\csc^2 x \).
  • Derivative of \( \csc x \): Related, \( \csc x \) has a derivative of \(-\csc x \cot x \), showcasing how derivatives in trigonometry intertwine.
In the original problem, the term \( x^2 \cot x \) required us to use this trigonometric derivative alongside the product rule. So, the added complexity of the problem stems in part from the nature of \( \cot x \) and its derivative.
Understanding these derivatives enables calculus students to handle a wide variety of problems involving sine, cosine, tangent, and other trigonometric functions.

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Most popular questions from this chapter

The linearization is the best linear approximation This is why we use the linearization.) Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the erroo \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=\) \(f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

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