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Chapter 27: Problem 15

Find the derivatives of the given functions. $$y=\sqrt{\sec 4 x}$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = 2\sec^{-1/2}(4x)\tan(4x) \).

Step by step solution

01

Identify the Outer Function

The given function is \( y = \sqrt{\sec 4x} \). This is a composition of functions, where the outer function is \( \sqrt{u} = u^{1/2} \).
02

Differentiate the Outer Function

To differentiate \( u^{1/2} \), use the power rule: \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \). Thus, the derivative of the outer function with respect to \( u \) is \( \frac{1}{2}\sec^{1/2}(4x) \cdot (\sec(4x))^{-1/2} \).
03

Identify the Inner Function

Here, the inner function is \( u = \sec(4x) \). Our task is to differentiate this with respect to \( x \).
04

Differentiate the Inner Function

The derivative of \( \sec(u) \) is \( \sec(u)\tan(u) \). Further chain it with the derivative of \( 4x \). For \( u = \sec(4x) \), the derivative is \( \sec(4x)\tan(4x) \cdot 4 \).
05

Apply the Chain Rule

Multiply the derivative of the outer function by the derivative of the inner function according to the chain rule: \( \frac{dy}{dx} = \frac{1}{2}\sec^{1/2}(4x) \cdot (\sec(4x))^{-1/2} \cdot \sec(4x)\tan(4x) \cdot 4 \).
06

Simplify the Expression

Combine and simplify the expression. That yields \( \frac{dy}{dx} = 2\sec^{-1/2}(4x)\tan(4x) \).

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Key Concepts

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

Composition of Functions
In mathematics, a composition of functions involves inserting one function inside another. Think of it as stacking two functions on top of each other. In our case, the given function is
  • The outer function: \( y = \sqrt{u} \), which can also be written in the form \( y = u^{1/2} \).
  • The inner function: \( u = \sec(4x) \).
When you deal with compositions of functions, the key is to recognize which function is outside and which one is inside. This concept is foundational for understanding how the chain rule works, as you will be calculating derivatives of these functions accordingly. By separating functions into their outer and inner parts, it becomes easier to apply derivative rules systematically.
Power Rule
The power rule is a straightforward and one of the most frequently used rules in calculus for finding derivatives. It states that for any function of the form \( u^{n} \), the derivative is given by multiplying the power \( n \) by the base \( u \) raised to the power of \( n-1 \). For example,
  • \( \frac{d}{du}(u^{n}) = nu^{n-1} \)
  • In our case, when differentiating \( y = \sqrt{u} = u^{1/2} \), we used the power rule to find \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \).
So for any function involving a power, the power rule enables a simple way to find its rate of change. It's essential to master this rule to handle more complex differentiation problems effectively.
Chain Rule
The chain rule is a fundamental principle used to find the derivative of a composition of functions. It articulates how to interrelate the rates of change of the inner and outer functions. The basic form of the chain rule might be expressed as follows:
  • If \( y = f(g(x)) \), then its derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
  • Using this in our problem: identify the inner function as \( u = \sec(4x) \) and its derivative as \( \sec(4x)\tan(4x) \cdot 4 \). The outer function's derivative is \( \frac{1}{2}u^{-1/2} \).
By following the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. This allows us to find the complete derivative of our initial composite function \( y = \sqrt{\sec 4x} \).
Trigonometric Functions
Trigonometric functions include sine, cosine, tangent, and their reciprocals such as secant, cosecant, and cotangent. In calculus, understanding how to differentiate these functions is crucial. Here's a quick rundown:
  • \( \frac{d}{dx} (\sec x) = \sec x \tan x \)
  • \( \frac{d}{dx} (\tan x) = \sec^2 x \)
For the given problem, we focused on \( \sec(4x) \). We first found its derivative as \( \sec(4x)\tan(4x) \times 4 \). With trigonometric derivatives, one common approach is to first recognize the function and use known derivative results, then apply them to solve more complicated problems as seen in this exercise.

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Most popular questions from this chapter

Solve the given problems. Display the graph of \(y=e^{x}\) on a calculator. Using the derivative feature, evaluate \(d y / d x\) for \(x=2\) and compare with the value of \(y\) for \(x=2\).

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.30 shows the graph of \(y=\sinh x\). The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{u}+e^{-u}\right)\) Figure 27.31 shows the graph of \(y=\cosh x\). These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\). Show that \(\frac{d^{2} \sinh x}{d x^{2}}=\sinh x \quad\) and \(\quad \frac{d^{2} \cosh x}{d x^{2}}=\cosh x\).

Solve the given problems by finding the appropriate derivative. Assuming that force is proportional to acceleration, show that a particle moving along the \(x\) -axis, so that its displacement \(x=a e^{k t}+b e^{-k t},\) has a force acting on it which is proportional to its displacement.

Solve the given problems by finding the appropriate derivative. Find the values of \(x\) for which the graphs of \(y_{1}=e^{2 / x^{2}}\) and \(y_{2}=e^{-2 / x^{2}}\) are increasing and decreasing. Explain why they differ as they do.

Solve the given problems. For what values of \(m\) does the function \(y=a e^{m x}\) satisfy the equation \(y^{\prime \prime}+y^{\prime}-6 y=0 ?\)
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