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Chapter 3: Problem 32

In Exercises \(27-32,\) find \(d p / d q\) $$ p=\frac{3 q+\tan q}{q \sec q} $$

Short Answer

Expert verified
Apply the quotient rule, then simplify to find the derivative.

Step by step solution

01

Differentiate Numerator and Denominator

First, identify the fraction: the numerator is \(3q + \tan q\) and the denominator is \(q \sec q\). Differentiate the numerator separately for each term: the derivative of \(3q\) is 3 and the derivative of \(\tan q\) is \(\sec^2 q\). Now, differentiate the denominator: \(q \sec q\) gives \(\sec q + q \sec q \tan q\) using the product rule.
02

Apply the Quotient Rule

Use the quotient rule for differentiation, which states \( \frac{d}{dq} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dq} - u \cdot \frac{dv}{dq}}{v^2} \). Here, \(u = 3q + \tan q\) and \(v = q \sec q\). Substitute the derivatives: \[ dp/dq = \frac{(q \sec q) \cdot (3 + \sec^2 q) - (3q + \tan q) \cdot (\sec q + q \sec q \tan q)}{(q \sec q)^2} \]
03

Simplify the Expression

Expand and simplify the expression. Multiply out terms in the numerator: \((q \sec q) \cdot (3 + \sec^2 q) = 3q \sec q + q \sec^3 q\) and \((3q + \tan q) \cdot (\sec q + q \sec q \tan q)\). Simplify each part and combine like terms to clean up the expression.
04

Final Calculation for \(d p/d q\)

Combine all simplified expressions from both the numerator and the denominator. Make sure to account for negative and positive signs. Once simplified completely, rewrite the expression to find \(\frac{dp}{dq}\). The final expression will be a combination of terms from the expanded form.

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Key Concepts

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

Derivative Calculation
Calculating derivatives is a fundamental skill in calculus, as it helps in understanding how functions change. The derivative of a function at any point gives us the slope of the tangent to the function's graph at that point. When faced with a function consisting of a fraction like \(p = \frac{3q + \tan q}{q \sec q}\), it's essential to apply the right differentiation rules to each part.

To find the derivative \( \frac{dp}{dq} \) of the given function, we must first identify and differentiate both the numerator \((3q + \tan q)\) and the denominator \((q \sec q)\). Here's a step-by-step approach:
  • Differentiate the numerator: The derivative of \(3q\) is \(3\), while the derivative of \(\tan q\) is \(\sec^2 q\) because of trigonometric derivative rules.
  • Differentiate the denominator: For \(q \sec q\), use the product rule to get \(\sec q + q \sec q \tan q\).
Once both derivatives are found, we apply the quotient rule to combine them and find \( \frac{dp}{dq} \). Using these rules allows us to handle complex functions with various terms.
Product Rule
The product rule is used in calculus when differentiating expressions that are products of two functions. The primary idea is to take the derivative of both functions separately, sum their products, and then add them together.

When dealing with \(q \sec q\) in the denominator of our function, we apply the product rule. The product rule formula is \( \frac{d}{dq}(f(q)g(q)) = f'(q)g(q) + f(q)g'(q) \). Here's how it works in this specific case:
  • Let \(f(q) = q\) and \(g(q) = \sec q\).
  • The derivative \(f'(q) = 1\) and \(g'(q) = \sec q \tan q\).
  • Apply the product rule: The result is \(1 \cdot \sec q + q \cdot \sec q \tan q\).
Understanding and using the product rule correctly is crucial for solving problems involving products of functions, ensuring accuracy in derivative calculations.
Trigonometric Derivatives
Trigonometric derivatives are essential when differentiating functions involving trigonometric expressions. Knowing the derivatives of basic trigonometric functions helps simplify complex calculations.

In the exercise, we encounter \(\tan q\) and \(\sec q\), which have their specific derivatives:
  • The derivative of \(\tan q\) is \(\sec^2 q\). This comes from understanding that each trigonometric function has a corresponding derivative linked to another trigonometric function.
  • For \(\sec q\), the derivative is \(\sec q \tan q\). Knowing these derivatives allows one to manipulate and solve problems involving trigonometric expressions effectively.
By applying trigonometric derivatives along with other differentiation rules, students can tackle a broad range of calculus problems involving trigonometric functions.

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Most popular questions from this chapter

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} y^{3}+\cos x y=x^{2}, \quad P(1,0) \end{equation}

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2} $$

Quadratic approximations $$ \begin{array}{l}{\text { a. Let } Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2} \text { be a quadratic }} \\ {\text { approximation to } f(x) \text { at } x=a \text { with the properties: }}\end{array} $$ $$ \begin{aligned} \text { i) } Q(a) &=f(a) \\ \text { ii) } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { iii) } Q^{\prime \prime}(a) &=f^{\prime \prime}(a) \end{aligned} $$ $$ \text{Determine the coefficients}b_{0}, b_{1}, \text { and } b_{2} $$ $$ \begin{array}{l}{\text { b. Find the quadratic approximation to } f(x)=1 /(1-x) \text { at }} \\ {\quad x=0 .} \\ {\text { c. } \operatorname{Graph} f(x)=1 /(1-x) \text { and its quadratic approximation at }} \\ {x=0 . \text { Then zoom in on the two graphs at the point }(0,1) \text { . }} \\ {\text { Comment on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { d. Find the quadratic approximation to } g(x)=1 / x \text { at } x=1} \\ {\text { Graph } g \text { and its quadratic approximation together. Comment }} \\ {\text { on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { e. Find the quadratic approximation to } h(x)=\sqrt{1+x} \text { at }} \\ {x=0 . \text { Graph } h \text { and its quadratic approximation together. }} \\ {\text { Comment on what you see. }} \\\ {\text { f. What are the linearizations of } f, g, \text { and } h \text { at the respective }} \\ {\text { points in parts (b), (d), and (e)? }}\end{array} $$

If \(x^{2} y^{3}=4 / 27\) and \(d y / d t=1 / 2,\) then what is \(d x / d t\) when \(x=2 ?\)

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