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Chapter 2: Problem 31

Find \(d y / d x\) \(y=\cos (\cos x)\)

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = \sin(\cos x) \cdot \sin x \).

Step by step solution

01

Understand the Function Composition

The function we need to differentiate is a composite function: \( y = \cos(\cos x) \). It involves the composition of two cosine functions.
02

Apply the Chain Rule

Since \( y = \cos u \) where \( u = \cos x \), we'll apply the chain rule. The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
03

Differentiate Outer Function

First, differentiate the outer function \( \cos u \) with respect to \( u \). The derivative of \( \cos u \) is \(-\sin u \). Thus, \( \frac{dy}{du} = -\sin(\cos x) \).
04

Differentiate Inner Function

Now, differentiate the inner function \( \cos x \) with respect to \( x \). The derivative is \(-\sin x \). So, \( \frac{du}{dx} = -\sin x \).
05

Calculate \(\frac{dy}{dx}\)

Use the chain rule to find \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute \(-\sin(\cos x)\) for \( \frac{dy}{du} \) and \(-\sin x\) for \( \frac{du}{dx} \):\[ \frac{dy}{dx} = -\sin(\cos x) \times (-\sin x) \] Simplify to get \[ \frac{dy}{dx} = \sin(\cos x) \cdot \sin x \].

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It essentially describes how to calculate the derivative of a function composed of other functions. If you have a function that is the composition of two or more functions, the chain rule is your go-to tool.To apply the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. In mathematical terms, if you have a composite function like \( y = f(g(x)) \), the derivative, \( \frac{dy}{dx} \), is calculated as:
  • First, find the derivative of the outer function, \( f \), with respect to the inner function \( g \). This gives \( \frac{df}{dg} \).
  • Next, find the derivative of the inner function, \( g \), with respect to \( x \), resulting in \( \frac{dg}{dx} \).
  • Finally, multiply these derivatives together: \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \).
This ensures all parts of the composite function are properly accounted for in the differentiation process.
Composite Functions
Composite functions involve one function nested within another. Understanding how these functions work is essential for applying the chain rule effectively.A composite function can be viewed as a function applied to the results of another function. For example, if you have \( y = \cos(\cos x) \), this describes a situation where the cosine function is applied to the results of another cosine function.
  • The inner function here is \( g(x) = \cos x \).
  • The outer function is \( f(u) = \cos u \), where \( u = g(x) \).
When differentiating, always identify the inner and outer functions. Knowing them helps in applying the chain rule correctly. You'll first differentiate the outer function while keeping the inner function unchanged, then differentiate the inner function.
Trigonometric Derivatives
Trigonometric derivatives are specific rules for differentiating trigonometric functions. These are essential when dealing with calculus problems involving trigonometric functions, like sine and cosine.The basic derivatives you'll often use are:
  • For the sine function, the derivative of \( \sin x \) is \( \cos x \).
  • For the cosine function, the derivative of \( \cos x \) is \( -\sin x \).
In our original exercise with \( y = \cos(\cos x) \), both the inner and outer functions involved are cosine functions. So, knowing that the derivative of \( \cos(u) \) is \( -\sin(u) \) was crucial.
  • We differentiate the outer function first, giving \( -\sin(\cos x) \).
  • We then differentiate the inner function, \( \cos x \), resulting in \( -\sin x \).
Finally, these trigonometric derivatives are combined through the chain rule to compute the full derivative of the composite function.

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Most popular questions from this chapter

The force \(F\) (in pounds) acting at an angle \(\theta\) with the horizontal that is needed to drag a crate weighing \(W\) pounds along a horizontal surface at a constant velocity is given by $$ F=\frac{\mu W}{\cos \theta+\mu \sin \theta} $$ where \(\mu\) is a constant called the coefficient of sliding friction between the crate and the surface (see the accompanying figure). Suppose that the crate weighs \(150 \mathrm{lb}\) and that \(\mu=0.3\) (a) Find \(d F / d \theta\) when \(\theta=30^{\circ} .\) Express the answer in units of pounds/degree. (b) Find \(d F / d t\) when \(\theta=30^{\circ}\) if \(\theta\) is decreasing at the rate of \(0.5^{\circ} / \mathrm{s}\) at this instant.

Determine whether the statement is true or false. Explain your answer. If \(f(x)\) is a cubic polynomial, then \(f^{\prime}(x)\) is a quadratic polynomial.

In each part, compute \(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}\), and then state the formula for \(f^{(n)}\). (a) \(f(x)=1 / x\) (b) \(f(x)=1 / x^{2}\) [Hint: The expression \((-1)^{n}\) has a value of 1 if \(n\) is even and \(-1\) if \(n\) is odd. Use this expression in your answer.]

The "co" in "cosine" comes from "complementary," since the cosine of an angle is the sine of the complementary angle, and vice versa: $$ \cos x=\sin \left(\frac{\pi}{2}-x\right) \text { and } \sin x=\cos \left(\frac{\pi}{2}-x\right) $$ Suppose that we define a function \(g\) to be a cofunction of a function \(f\) if $$ g(x)=f\left(\frac{\pi}{2}-x\right) \quad \text { for all } x $$ Thus, cosine and sine are cofunctions of each other, as are cotangent and tangent, and also cosecant and secant. If \(g\) is the cofunction of \(f\), state a formula that relates \(g^{\prime}\) and the cofunction of \(f^{\prime}\). Discuss how this relationship is exhibited by the derivatives of the cosine, cotangent, and cosecant functions.

Find an equation for the tangent line to the graph at the specified value of \(x\). \(y=3 \cot ^{4} x, x=\frac{\pi}{4}\)
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