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Chapter 3: Problem 6

In Exercises \(1-18,\) 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 the first term

The first term of the function is \(x^2 \cot x\). We need to use the product rule to differentiate this. The product rule states \((uv)' = u'v + uv'\). Here, \(u = x^2\) and \(v = \cot x\). 1. Compute \(u' = \frac{d}{dx}(x^2) = 2x\).2. Compute \(v' = \frac{d}{dx}(\cot x) = -\csc^2 x\).3. Apply the product rule: \[\frac{d}{dx}(x^2 \cot x) = (2x)(\cot x) + (x^2)(-\csc^2 x) = 2x\cot x - x^2\csc^2 x.\]
02

Differentiate the second term

The second term of the function is \(-\frac{1}{x^2}\). To differentiate this, write it as \(-x^{-2}\). Then, apply the power rule:\[ \frac{d}{dx}(-x^{-2}) = -(-2)x^{-3} = 2x^{-3}.\]
03

Combine the derivatives

Now that we have the derivatives of both terms, combine them:\[ \frac{dy}{dx} = (2x\cot x - x^2\csc^2 x) + 2x^{-3}.\]
04

Simplify the expression

Simplify the expression by combining like terms (if possible). Here, the expression is already simplified.Final derivative expression:\[ \frac{dy}{dx} = 2x\cot x - x^2\csc^2 x + \frac{2}{x^3}.\]

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Key Concepts

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

Product Rule
The product rule is a key concept in calculus. It helps us differentiate products of two functions. When two functions are multiplied together, their derivative isn't just the derivative of each function separately. Instead, we use the product rule:
  • If you have functions \( u \) and \( v \), their product is \( uv \). The derivative \((uv)'\) is given by \(u'v + uv'\).
  • Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.
This rule accounts for how both functions change.
In the original exercise, we differentiated \( x^2 \cot x \) by letting \( u = x^2 \) and \( v = \cot x \). By computing \( u' \) and \( v' \), and then applying the rule, we elegantly found the derivative of the product term.
Trigonometric Differentiation
Trigonometric differentiation involves finding derivatives of trigonometric functions, such as sine, cosine, and tangent. These functions have specific derivatives. Learning these is crucial. For example, the derivative of \( \cot x \) is \( -\csc^2 x \).
Here's a list of derivative formulas for common trig functions:
  • \(\frac{d}{dx} (\sin x) = \cos x\)
  • \(\frac{d}{dx} (\cos x) = -\sin x\)
  • \(\frac{d}{dx} (\tan x) = \sec^2 x\)
  • \(\frac{d}{dx} (\cot x) = -\csc^2 x\)
  • \(\frac{d}{dx} (\sec x) = \sec x \tan x\)
  • \(\frac{d}{dx} (\csc x) = -\csc x \cot x\)
In our exercise, this knowledge was applied to find \( v' \), the derivative of \( \cot x \), which is crucial for using the product rule.
Power Rule
The power rule is one of the simplest and most used rules in calculus. It tells us how to differentiate functions with the form \( x^n \). The rule is:
  • If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
For negative exponents, the rule still applies. You simply move the exponent and reduce it by one.
In the original solution, to differentiate \(-\frac{1}{x^2}\), we rewrote it as \(-x^{-2}\) and then used the power rule to find its derivative as \(2x^{-3}\).
Mastering this rule will greatly simplify your differentiation tasks and help you tackle more complex functions efficiently.

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Most popular questions from this chapter

Cylinder pressure If gas in a cylinder is maintained at a constant temperature \(T\) , the pressure \(P\) is related to the volume \(V\) by a formula of the form $$P=\frac{n R T}{V-n b}-\frac{a n^{2}}{V^{2}}$$ in which \(a, b, n,\) and \(R\) are constants. Find \(d P / d V .\) (See accompanying figure.)

A melting ice layer A spherical iron ball 8 in. in diameter is conted with a layer of ice of uniform thickness. If the ice melts at the rate of 10 \(\mathrm{in}^{3} / \mathrm{min}\) , how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?

A cube's surface area increases at the rate of 72 \(\mathrm{in}^{2} / \mathrm{sec} .\) At what rate is the cube's volume changing when the edge length is \(x=3\) in?

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the volume } V=\pi r^{2} h \text { of a right circular cylinder }} \\ {\text { when the radius changes from } r_{0} \text { to } r_{0}+d r \text { and the height does }} \\ {\text { not change }}\end{array} $$

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=\frac{x-1}{3 x^{2}+1}, \quad x_{0}=-1$$
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