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Chapter 3: Problem 33

Find the derivatives of the functions. $$f(x)=\sqrt{7+x \sec x}$$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{\sec x (1 + x \tan x)}{2\sqrt{7 + x \sec x}} \)."

Step by step solution

01

Use Derivative Rules

The function given is \( f(x) = \sqrt{7 + x \sec x} \). This function is a composition of two functions. The outer function is a square root, and the inner function is \( 7 + x \sec x \). To find the derivative, we will need to apply the chain rule. The chain rule states that \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).
02

Differentiate the Outer Function

The outer function is \( g(u) = \sqrt{u} \), where \( u = 7 + x \sec x \). The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \). Substituting back \( u = 7 + x \sec x \), we have \( \frac{1}{2\sqrt{7 + x \sec x}} \).
03

Differentiate the Inner Function

The inner function is \( h(x) = 7 + x \sec x \). The derivative of \( 7 \) is \( 0 \), and the derivative of \( x \sec x \) requires the product rule: \( (uv)' = u'v + uv' \). Let \( u = x \) and \( v = \sec x \). Then \( u' = 1 \) and \( v' = \sec x \tan x \). Applying the product rule, \( \frac{d}{dx}(x \sec x) = 1 \cdot \sec x + x \cdot \sec x \tan x \). Thus, \( h'(x) = \sec x + x \sec x \tan x \).
04

Apply the Chain Rule

Use the chain rule to combine the derivatives from Step 2 and Step 3: \( f'(x) = \frac{1}{2\sqrt{7 + x \sec x}} \cdot (\sec x + x \sec x \tan x) \).
05

Simplify the Expression

The derivative of \( f(x) = \sqrt{7 + x \sec x} \) is: \( f'(x) = \frac{\sec x (1 + x \tan x)}{2\sqrt{7 + x \sec x}} \).

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Key Concepts

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

Understanding the Chain Rule
When working with derivatives, especially for functions that are compositions of other functions, the chain rule is your best friend. Suppose you have a function like \( f(g(x)) \). The chain rule helps you find the derivative by first taking the derivative of the outer function \( f \), keeping the inner function \( g(x) \) intact. Then, you multiply it by the derivative of the inner function.

So, mathematically, in simple terms:
  • First, find \( f'(g(x)) \), which is the derivative of the outer function.
  • Then, find \( g'(x) \), which is the derivative of the inner function.
  • Multiply the two results together: \( f'(g(x)) \cdot g'(x) \).
Using the chain rule makes it possible to handle complex functions by breaking them into simpler parts, making derivative calculus much more manageable.
Mastering the Product Rule
In calculus, whenever you differentiate functions that are products of two separate parts, the product rule comes into play.

Suppose you have a function represented as \( u(x) \cdot v(x) \), where both \( u \) and \( v \) depend on \( x \). The product rule states:
  • Differentiate \( u \) to get \( u' \).
  • Differentiate \( v \) to get \( v' \).
  • The derivative of the product is \( u'v + uv' \).
This rule simplifies tackling derivatives of products, ensuring all factors are appropriately accounted for in your calculations. Once you familiarize yourself with the product rule, finding derivatives of multiplied functions becomes a straightforward task.
Calculating Trigonometric Derivatives
Trigonometric functions, like \( \sin x \), \( \cos x \), and \( \sec x \), have specific rules for their derivatives. Knowing these derivative rules by heart simplifies your work.

For example:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \sec x \) is \( \sec x \tan x \).
These rules allow you to efficiently differentiate trigonometric functions, whether on their own or when combined with other functions. In our original problem, we used the derivative for \( \sec x \) to handle part of the inner function, demonstrating just how essential this knowledge is for solving calculus problems with trigonometric components.

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Most popular questions from this chapter

What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$f(h)=\frac{\sin h}{h}$$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$\lim _{h \rightarrow 0} \frac{\cos h-1}{h}$$ c. Now go back to the derivation of the formula for the derivative of \(\sin x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of \(\cos x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of \(\sin x\) and \(\cos x ?\)

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

Show that if it is possible to draw three normals from the point \((a, 0)\) to the parabola \(x=y^{2}\) shown in the accompanying diagram, then \(a\) must be greater than \(1 / 2\). One of the normals is the \(x\) -axis. For what value of \(a\) are the other two normals perpendicular?

Find \(d y / d t\) $$y=\left(t^{-3 / 4} \sin t\right)^{4 / 3}$$

Find \(d y / d t\) $$y=(t \tan t)^{10}$$
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