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Chapter 3: Problem 5

Given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\) $$y=\sqrt{u}, \quad u=\sin x$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{\cos x}{2\sqrt{\sin x}} \)

Step by step solution

01

Identify the Functions

First, identify the given functions from the problem statement. We have \( y = \sqrt{u} \) and \( u = \sin x \).
02

Derivative of Inner Function

Find the derivative of \( u \) with respect to \( x \), denoted as \( g'(x) \). Since \( u = \sin x \), we have \( g'(x) = \cos x \).
03

Derivative of Outer Function

Find the derivative of \( y \) with respect to \( u \), denoted as \( f'(u) \). Since \( y = \sqrt{u} = u^{1/2} \), using the power rule, we get \( f'(u) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \).
04

Apply the Chain Rule

Use the chain rule to find \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Substitute \( g(x) = \sin x \) and \( g'(x) = \cos x \) into the equation. This gives \( \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x}} \cdot \cos x \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It is all about finding how a function changes as its input changes. Imagine you're driving a car and the speedometer shows how fast you're going at every moment. Differentiation works like that speedometer, measuring the rate of change of a function.

Imagine you have a function, say \(y\), that depends on \(x\). To differentiate \(y\) with respect to \(x\), you denote this rate of change as \( \frac{dy}{dx} \). This tells you how much \(y\) changes when \(x\) changes by a little bit.
  • The process often involves rules like the power rule, product rule, and chain rule, which help simplify the differentiation process.
  • By knowing these rules, you can tackle complex functions by breaking them down into simpler parts.
Composite Functions
Composite functions are combinations of two functions where the output of one function becomes the input of another. This is like a factory line, where one machine processes a part and then passes it to the next machine for further processing.

In mathematical terms, if you have two functions \(f\) and \(g\), creating a composite function \(f(g(x))\) means you first apply \(g\) to \(x\), and then apply \(f\) to the result of \(g(x)\). Here, the function \(f\) operates on the results that come from \(g\).
  • This concept is key when dealing with functions within functions, like \(y = \sqrt{u}\) where \(u = \sin x\).
  • Using correct substitution and differentiation techniques can help solve such expressions efficiently.
Derivatives of Trigonometric Functions
Trigonometric functions are a staple in calculus. They describe how angles relate to sides in triangles, yet they often pop up in wave patterns and oscillations. Key trig functions include sine, cosine, and tangent.

When it comes to differentiation, each trigonometric function has a unique derivative:
  • The derivative of \(\sin x\) is \(\cos x\).
  • The derivative of \(\cos x\) is \(-\sin x\).
  • The derivative of \(\tan x\) might involve a bit more complexity, leading to \(\sec^2 x\).

These derivatives are crucial because they allow us to find the rates of change of trigonometric functions concerning angles or time. Whether you are an engineer or a scientist, understanding these derivatives helps in modeling real-world phenomena.

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Most popular questions from this chapter

Find \(y^{\prime \prime}\) $$y=9 \tan \left(\frac{x}{3}\right)$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L\). In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2\)

The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to \(\sqrt{s}\) when it is \(s\) km from Earth's center. Show that the meteorite's acceleration is inversely proportional to \(s^{2}\)

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. $$f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1$$
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