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Chapter 3: Problem 18

find \(d y\). \(y=x \sqrt{1-x^{2}}\)

Short Answer

Expert verified
The differential is \(dy = \frac{1-x^2}{\sqrt{1-x^2}}dx\).

Step by step solution

01

Identify the Functions

The given function is a product: \(y = x\sqrt{1-x^2}\). Here, \(x\) is the first function and \(\sqrt{1-x^2}\) is the second function.
02

Use the Product Rule

To find \(dy\), we need to differentiate \(y\) using the product rule. The product rule is \(uv = u'v + uv'\) where \(u\) and \(v\) are functions of \(x\). Hence, \(dy = (1)\sqrt{1-x^2} + x\left(\frac{d}{dx}\sqrt{1-x^2}\right)dx\).
03

Differentiate the Square Root Function

Differentiate \(\sqrt{1 - x^2}\) with respect to \(x\). Use the chain rule: \(\frac{d}{dx}\sqrt{1-x^2} = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}\).
04

Substitute Back Into the Product Rule

Substitute the derivative of the second function from Step 3 into the product rule: \(dy = \sqrt{1-x^2}dx + x\left(\frac{-x}{\sqrt{1-x^2}}\right)dx\).
05

Simplify the Expression

Combine the terms: \(dy = \sqrt{1-x^2}dx - \frac{x^2}{\sqrt{1-x^2}}dx = \left(\frac{1-x^2}{\sqrt{1-x^2}}\right)dx\).
06

Final Expression for dy

Simplify the expression: \(dy = \frac{1-x^2}{\sqrt{1-x^2}}dx\).

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Key Concepts

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

Differentiation
Differentiation is one of the cornerstones of calculus, allowing us to understand how a function's value changes with respect to changes in its input. In essence, differentiation provides us with the derivative, a fundamental tool that represents the slope or rate of change of a function at any given point. Let's break this down:
  • The derivative of a function tells us how the function's output changes as the input changes in infinitesimal amounts.
  • Mathematically, if we have a function \( f(x) \), its derivative is represented as \( f'(x) \) or \( \frac{df}{dx} \).
  • Differentiation rules, like the product rule and the chain rule, guide us in finding derivatives of complex expressions.
Using the product rule for differentiation, as in the step-by-step solution provided, helps in managing functions that are products of two or more simpler functions.
Chain Rule
The chain rule is a crucial method for differentiating composite functions, where one function is nested inside another. In simple terms, when you have a function within a function, the chain rule is your go-to. Here鈥檚 how it works and why it鈥檚 indispensable:
  • The chain rule states that if a variable \( y \) is a function of \( u \), which is itself a function of \( x \), then the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
  • In practice, this means taking the derivative of the outer function as it is and multiplying it by the derivative of the inner function.
  • The chain rule helps us differentiate complicated expressions such as \( \sqrt{1 - x^2} \), as seen in the solution where its derivative is found as \( \frac{-x}{\sqrt{1-x^2}} \).
Understanding the chain rule expands your capability to handle diverse and more complex functions.
Calculus Concepts
Calculus is a branch of mathematics that involves the study of rates of change and the accumulation of quantities. Two main concepts in calculus are differentiation and integration, each serving unique purposes:
  • Differentiation: As explored earlier, this helps determine the rate at which function values change, providing tools like slopes of tangents and velocities.
  • Integration: While differentiation focuses on rates of change, integration deals with the accumulation of quantities, like calculating areas under curves.
Calculus concepts apply widely across fields such as physics, engineering, economics, and biology, helping to solve problems involving dynamic systems. Grasping these concepts, therefore, equips students with a powerful toolkit for analyzing and understanding a vast range of real-world phenomena.

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Most popular questions from this chapter

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1\) $$\begin{array}{|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 0 & {1} & {1} & {5} & {1 / 3} \\ \hline 1 & {3} & {-4} & {-1 / 3} & {-8 / 3} \\ \hline\end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x .\) $$\begin{array}{l}{\text { a. } 5 f(x)-g(x), \quad x=1 \quad \text { b. } f(x) g^{3}(x), \quad x=0}\end{array}$$ $$ $$$$c .\frac{f(x)}{g(x)+1}, \quad x=1 \quad \text { d. } f(g(x)), \quad x=0$$ $$ \begin{array}{l}{\text { e. } g(f(x)), \quad x=0 \quad \text { f. }\left(x^{11}+f(x)\right)^{-2}, \quad x=1} \\ {\text { g. } f(x+g(x)), \quad x=0}\end{array} $$

Commercial air traffic Two commercial airplanes are flying at an altitude of \(40,000\) ft along straight-line courses that intersect at right angles. Plane \(A\) is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd \() .\) Plane \(B\) is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when \(A\) is 5 nautical miles from the intersection point and \(B\) is 12 nautical miles from the intersection point?

Consider the function $$f(x)=\left\\{\begin{array}{cc}{x \sin \left(\frac{1}{x}\right),} & {x>0} \\\ {0,} & {x \leq 0}\end{array}\right.$$ a. Show that \(f\) is continuous at \(x=0\) b. Determine \(f^{\prime}\) for \(x \neq 0\) . c. Show that \(f\) is not differentiable at \(x=0\) .

Graph \(y=-\sin x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$ y=\frac{\cos (x+h)-\cos x}{h} $$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+} ?\) As \(h \rightarrow 0^{-9}\) What phenomenon is being illustrated here?

Graph \(y=\cos x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$ y=\frac{\sin (x+h)-\sin x}{h} $$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+} ?\) As \(h \rightarrow 0^{-}\) ? What phenomenon is being illustrated here?
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