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Find the derivative of the function. $$ f(x)=\sin (\sin x) $$

Short Answer

Expert verified
The derivative of the function \( f(x) = \sin(\sin{x}) \) is \( f'(x) = \cos(\sin(x)) \cdot \cos(x) \).

Step by step solution

01

Find the derivative of the outer function

First, let's find the derivative of the outer function, \( g(x) = \sin(x) \). The derivative of the sine function is the cosine function, so we have \( g'(x) = \cos(x) \).
02

Find the derivative of the inner function

Now, let's find the derivative of the inner function, \( f(x) = \sin(x) \). As mentioned before, the derivative of the sine function is the cosine function, so we have \( f'(x) = \cos(x) \).
03

Apply the chain rule

Now that we have the derivatives of both the inner and the outer functions, we can apply the chain rule to find the derivative of the composite function. We have \( g'(f(x)) \cdot f'(x) = \cos(\sin(x)) \cdot \cos(x) \). Thus, the derivative of the given function is: \[ f'(x) = \cos(\sin(x)) \cdot \cos(x) \]

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Key Concepts

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

Derivative of Sine Function
The derivative of the sine function is a fundamental concept in calculus, essential for solving various types of problems. When we talk about the derivative of the sine function, \( \sin(x) \), the result is \( \cos(x) \). This means that if you have a function \( f(x) = \sin(x) \), the rate at which \( f(x) \) changes with respect to \( x \) is given by \( f'(x) = \cos(x) \).

This is crucial because sine and cosine are periodic functions that depict wave-like behavior, often representing real-world phenomena such as sound waves or alternating current. Understanding how to differentiate these functions allows us to predict the behavior of such periodic processes. Remember, the derivative is all about finding the rate of change, and with trigonometric functions like sine, we're often seeking to understand how quickly these waves are oscillating as time progresses.
Composite Function Derivatives
In calculus, a composite function is a function made up of two or more other functions. When you want to find the derivative of a composite function, you’ll often need to use the chain rule. To break this down, let's consider two functions, \( g(x) \) and \( f(x) \), where the composite function is \( h(x) = g(f(x)) \). The derivative of this composite function requires us to find the derivatives of \( g(x) \) and \( f(x) \) individually first.

The concept of derivatives of composite functions extends beyond just an academic exercise; it has practical applications in fields such as economics, engineering, and the sciences. Whenever one quantity depends on another, which in turn depends on a third variable, composite function derivatives come into play. For example, it could help us understand how the pressure of a gas changes with volume when the volume itself is changing with time.
Applying the Chain Rule in Calculus
The chain rule is a powerful tool in calculus, allowing us to differentiate composite functions. It essentially tells us how to take the derivative of a function nested within another function. To apply the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. In mathematical terms, if \( h(x) = g(f(x)) \) is a composite function, then the derivative \( h'(x) \) is \( g'(f(x)) \cdot f'(x) \).

For instance, if you need to differentiate \( h(x) = \sin(\sin x) \), you would find that \( h'(x) = \cos(\sin x) \cdot \cos(x) \) by applying the chain rule. This concept not only applies to trigonometric functions but to any set of functions that can be combined. Grasping the chain rule is crucial for solving a plethora of problems in differential calculus and for understanding how different rates of change interact with each other.

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Most popular questions from this chapter

A division of Ditton Industries manufactures the "Spacemaker" model microwave oven. Suppose that the daily total cost (in dollars) of manufacturing \(x\) microwave ovens is $$C(x)=0.0002 x^{3}-0.06 x^{2}+120 x+6000$$ What is the marginal cost when \(x=200\) ? Compare the result with the actual cost incurred by the company in manufacturing the 201 st oven.

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves. In Exercises \(89-92\), show that the curves with the given equations are orthogonal.$$ x^{2}-y^{2}=3, \quad x y=2 $$

A horizontal uniform beam of length \(L\) is supported at both ends and bends under its own weight \(w\) per unit length. Because of its elasticity, the beam is distorted in shape, and the resulting distorted axis of symmetry (shown dashed in the figure) is called the elastic curve. It can be shown that an equation for the elastic curve is $$y=\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right)$$ where the product \(E I\) is a constant called the flexural rigidity. (a) The distorted beam (b) The elastic curve in the \(x y\) -plane (The positive direction of the \(y\) -axis is directed downward.) a. Find the angle that the elastic curve makes with the positive \(x\) -axis at each end of the beam in terms of \(w, E\), and \(I .\) b. Show that the angle that the elastic curve makes with the horizontal at \(x=L / 2\) is zero. c. Find the deflection of the beam at \(x=L / 2\). (We will show that the deflection is maximal in Section 3.1, Exercise 74.)

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=x^{5}+2 x^{3}+x-1 ; \quad(0,-1) $$

Find an equation of the tangent line to the given curve at the indicated point. $$ y^{2}-x y^{2}-x^{3}=0 ; \quad\left(\frac{1}{2}, \frac{1}{2}\right) $$

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