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In Exercises \(1-18,\) find \(d y / d x\) $$ f(x)=\sin x \tan x $$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = \cos x \tan x + \sin x \sec^2 x \).

Step by step solution

01

Identify the Differentiation Rule

The given function is a product of two functions, \(f(x) = \sin x \cdot \tan x\). To find \( \frac{dy}{dx} \), the derivative of \(f(x)\), we need to apply the product rule, which states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then \((uv)' = u'v + uv'\).
02

Define Functions and Derivatives

Define \(u(x) = \sin x\) and \(v(x) = \tan x\). Find their derivatives: \(u'(x) = \cos x\) and \(v'(x) = \sec^2 x\).
03

Apply the Product Rule

Using the product rule: \((uv)' = u'v + uv'\), substitute the values: \(u = \sin x\), \(u' = \cos x\), \(v = \tan x\), \(v' = \sec^2 x\).
04

Calculate Each Term Separately

Calculate the first term: \(u'v = \cos x \cdot \tan x\). Calculate the second term: \(uv' = \sin x \cdot \sec^2 x\).
05

Combine the Terms

Add the two terms together to find the derivative: \( \frac{dy}{dx} = \cos x \tan x + \sin x \sec^2 x\).

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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 fundamental concept in calculus. Understanding these derivatives is essential for solving a wide array of calculus problems. In our example, we have the trigonometric functions \( \sin x \) and \( \tan x \). To find their derivatives, we rely on basic trigonometric differentiation rules:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \tan x \) is \( \sec^2 x \).
These rules are key in calculus, as they allow us to differentiate more complex expressions that involve trigonometric functions. When these functions are part of a larger equation, like a product, it's essential to utilize these derivatives accurately. This ensures that the entire expression can be differentiated correctly using additional rules such as the product rule. Familiarity with these derivatives reduces complexity when stepping through problems, aiding in efficient problem-solving.
Calculus Exercises
Calculus exercises are a practical way to apply the theoretical foundations of differentiation and integration. By working through these exercises, students can reinforce their understanding of key concepts and improve their problem-solving skills. In this exercise, we focus on differentiating the function \( f(x) = \sin x \tan x \). This function is particularly interesting because it combines two trigonometric functions, requiring a thoughtful application of calculus rules.
When approaching calculus exercises, it's crucial to:
  • Understand the functions involved and their respective roles in the given expression.
  • Identify the appropriate differentiation rules to apply. For instance, the product rule is necessary here because two functions are multiplied together.
  • Break the problem into manageable steps, as you differentiate each component before combining the results.
Exercises like this one not only test mathematical knowledge but also enhance strategic thinking, as different techniques might be combined in a single problem.
Step-by-Step Differentiation
Step-by-step differentiation is a systematic approach used to find derivatives of complex functions. The process involves breaking down the function into simpler components, determining the derivative of each, and then combining these results according to differentiation rules.
Let's reconsider our function \( f(x) = \sin x \tan x \):
  • Start by identifying each individual function in the product: \( u(x) = \sin x \) and \( v(x) = \tan x \).
  • Find the derivatives of these functions: \( u'(x) = \cos x \) and \( v'(x) = \sec^2 x \).
  • Apply the product rule, which is \((uv)' = u'v + uv'\), involving substituting the identified functions into this formula.
  • Calculate each term separately: \( u'v = \cos x \cdot \tan x \) and \( uv' = \sin x \cdot \sec^2 x \).
  • Finally, combine the results to obtain the derivative: \( \frac{dy}{dx} = \cos x \tan x + \sin x \sec^2 x \).
These detailed steps ensure a thorough understanding of how each part contributes to the final derivative, offering a clear and logical pathway to the solution. Following a structured approach makes complex differentiation tasks manageable and helps solidify fundamental calculus concepts.

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Most popular questions from this chapter

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

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} $$f(x)=x^{3}+x^{2}-x, x_{0}=1$$

Ships Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

Surface area Suppose that the radius \(r\) and surface area \(S=4 \pi r^{2}\) of a sphere are differentiable functions of \(t\) . Write an equation that relates \(d S / d t\) to \(d r / d t\) .

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} x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right) \end{equation}

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