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In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$y=\sec (\tan x)$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \sec(\tan x) \tan(\tan x) \cdot \sec^2(x) \)

Step by step solution

01

Identify Inner and Outer Functions

The given function is \( y = \sec (\tan x) \). We need to express it in the form \( y = f(u) \) where \( u = g(x) \). In this case, the inner function is \( g(x) = \tan x \), and therefore, \( u = \tan x \). The outer function is \( f(u) = \sec(u) \). Hence, we can write \( y = f(u) = \sec(u) \) where \( u = g(x) = \tan x \).
02

Differentiate Outer Function

Differentiate the outer function \( y = \sec(u) \) with respect to \( u \). The derivative of \( \sec(u) \) with respect to \( u \) is \( f'(u) = \sec(u)\tan(u) \).
03

Differentiate Inner Function

Differentiate the inner function \( u = \tan x \) with respect to \( x \). The derivative of \( \tan x \) is \( g'(x) = \sec^2(x) \).
04

Apply Chain Rule

Using the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), we substitute \( \frac{dy}{du} = \sec(u)\tan(u) \) and \( \frac{du}{dx} = \sec^2(x) \). Thus, \( \frac{dy}{dx} = \sec(u)\tan(u) \cdot \sec^2(x) \).
05

Substitute Back \(u\) as \(g(x)\)

Substitute \( u = \tan x \) back into the derivative: \( \frac{dy}{dx} = \sec(\tan x) \tan(\tan x) \cdot \sec^2(x) \). This is the expression of \( \frac{dy}{dx} \) as a function of \( x \).

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Key Concepts

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

Derivative of Trigonometric Functions
The derivative of trigonometric functions is a crucial concept in calculus. Trigonometric functions such as sine, cosine, and tangent have specific derivatives that are used frequently in calculus problems. Understanding these derivatives is essential for finding the rate of change of these functions.

In this example, we deal with the function \( y = \sec(\tan x) \). The derivatives of the relevant trigonometric functions are:
  • The derivative of \( \tan x \) is \( \sec^2 x \).
  • The derivative of \( \sec u \) with respect to \( u \) is \( \sec u \tan u \).
These derivatives are foundational, as they provide the necessary steps in our differentiation process. For any complex trigonometric composition, knowing these basic derivatives allows us to apply the chain rule effectively. This knowledge lets us transition smoothly through differentiating each layer of the function.
Function Composition
Function composition occurs when one function is applied inside another. In more straightforward terms, it’s like stacking functions one inside the other. The given problem deals with function composition where \( y = \sec(\tan x) \). Here, there are two functions nested together:
  • The inner function \( g(x) = \tan x \).
  • The outer function \( f(u) = \sec u \).
By recognizing this composition, we can break down the problem into manageable parts for differentiation.

First, we identify each function's role and differentiate them separately. This separation into inner and outer functions is crucial for applying differentiation techniques effectively. Once the functions are identified, we can proceed with the differentiation methods such as the chain rule.
Calculus Differentiation
Calculus differentiation is the process of finding the derivative, a measure of how a function changes as its input changes. In this scenario, the main tool we use is the **chain rule**, a fundamental differentiation technique.

The chain rule is used when dealing with compositions of functions. Specifically, when you have \( y = f(g(x)) \), the chain rule formula is:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]In our example, \( y = \sec(\tan x) \), the chain rule works as follows:
  • First, differentiate the outer function: \( \frac{dy}{du} = \sec(u)\tan(u) \).
  • Second, differentiate the inner function: \( \frac{du}{dx} = \sec^2(x) \).
  • Finally, multiply these derivatives: \( \frac{dy}{dx} = (\sec(\tan x) \tan(\tan x)) \cdot \sec^2(x) \).
This approach makes it easier to tackle complex functions by breaking them down into simple, diffentiable components. Mastery of this technique is essential when analyzing changes in multi-layered functions.

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Most popular questions from this chapter

A draining hemispherical reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius \(13 \mathrm{m},\) shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) meters deep. a. At what rate is the water level changing when the water is 8 m deep? b. What is the radius \(r\) of the water's surface when the water is \(\quad y\) m deep? c. At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

Find \(d s / d t\) when \(\theta=3 \pi / 2\) if \(s=\cos \theta\) and \(d \theta / d t=5\)

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} $$f(x)=\sin 2 x, \quad x_{0}=\pi / 2$$

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

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}

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