Problem 46
Question
Find \(\frac{d y}{d u}, \frac{d u}{d x},\) and \(\frac{d y}{d x}\) $$ y=\sqrt{u} \text { and } u=x^{2}-1 $$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{du} = \frac{1}{2\sqrt{u}}, \frac{du}{dx} = 2x, \frac{dy}{dx} = \frac{x}{\sqrt{u}} \).
1Step 1: Find \( \frac{dy}{du} \)
Since \( y = \sqrt{u} \), we need to find the derivative of \( y \) with respect to \( u \). Using the power rule, \( \sqrt{u} = u^{1/2} \) and the derivative \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \). This simplifies to \( \frac{1}{2\sqrt{u}} \). Thus, \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \).
2Step 2: Find \( \frac{du}{dx} \)
Given \( u = x^2 - 1 \), we take the derivative of \( u \) with respect to \( x \). Using the power rule, the derivative of \( x^2 \) is \( 2x \) and the derivative of a constant \(-1\) is 0. Therefore, \( \frac{du}{dx} = 2x \).
3Step 3: Use the Chain Rule to Find \( \frac{dy}{dx} \)
To find the derivative \( \frac{dy}{dx} \), we use the chain rule which states \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]. We already have \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \) and \( \frac{du}{dx} = 2x \). Thus, \[ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \times 2x = \frac{x}{\sqrt{u}} \].
Key Concepts
DerivativeChain Rule ApplicationPower Rule
Derivative
In the world of calculus, the concept of a derivative is foundational. It represents the rate at which a function is changing at any given point. When we discuss the derivative of a function with respect to a specific variable, we're essentially asking how that function changes as the variable itself changes. This is often visualized as the slope of the tangent line to the function's graph at a given point.
The process of finding a derivative is called differentiation, and it involves using specific rules depending on the form of the function. For instance, when dealing with a power function such as a square root, like in our exercise where we have \( y = \sqrt{u} \), we rely on the power rule for differentiation. Once the function is rewritten as \( u^{1/2} \), the derivative can be calculated using this rule. Thus, the derivative conveys vital information about the behavior and characteristics of a function, providing insights into how it evolves.
The process of finding a derivative is called differentiation, and it involves using specific rules depending on the form of the function. For instance, when dealing with a power function such as a square root, like in our exercise where we have \( y = \sqrt{u} \), we rely on the power rule for differentiation. Once the function is rewritten as \( u^{1/2} \), the derivative can be calculated using this rule. Thus, the derivative conveys vital information about the behavior and characteristics of a function, providing insights into how it evolves.
Chain Rule Application
The chain rule is a critical tool in the differentiation toolkit, particularly useful when dealing with composite functions. A composite function is essentially a function which is nested within another function. In our example, we have two functions: \( y = \sqrt{u} \) and \( u = x^2 - 1 \). Each of these is dependent on a different variable.
To find \( \frac{dy}{dx} \), which represents the derivative of \( y \) with respect to \( x \), we apply the chain rule. The chain rule elegantly connects the derivatives of these nested functions: - First, differentiate \( y \) with respect to \( u \) to obtain \( \frac{dy}{du} \). - Then, differentiate \( u \) with respect to \( x \) to find \( \frac{du}{dx} \). - Finally, the chain rule combines these to give \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
This step-by-step approach breaks down the seemingly complex problem into manageable steps, revealing how changes in \( x \) affect \( y \) through the intermediary function \( u \).
To find \( \frac{dy}{dx} \), which represents the derivative of \( y \) with respect to \( x \), we apply the chain rule. The chain rule elegantly connects the derivatives of these nested functions: - First, differentiate \( y \) with respect to \( u \) to obtain \( \frac{dy}{du} \). - Then, differentiate \( u \) with respect to \( x \) to find \( \frac{du}{dx} \). - Finally, the chain rule combines these to give \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
This step-by-step approach breaks down the seemingly complex problem into manageable steps, revealing how changes in \( x \) affect \( y \) through the intermediary function \( u \).
Power Rule
The power rule is a straightforward yet powerful technique for differentiating functions of the form \( x^n \), where \( n \) is any real number. This rule states that when taking the derivative of \( x^n \), the result is \( nx^{n-1} \). It simplifies the process of finding derivatives for polynomial-type expressions.
Using our exercise as an example, to differentiate \( u = x^2 - 1 \), we treat each term separately. For the term \( x^2 \), the power rule yields a derivative of \( 2x^{2-1} = 2x \). The constant term \(-1\) differentiates to zero. Applying the power rule in such situations significantly simplifies the work involved in finding derivatives, especially when variables are raised to significant powers. This reliability and ease of use make it a valuable tool for calculus students.
Using our exercise as an example, to differentiate \( u = x^2 - 1 \), we treat each term separately. For the term \( x^2 \), the power rule yields a derivative of \( 2x^{2-1} = 2x \). The constant term \(-1\) differentiates to zero. Applying the power rule in such situations significantly simplifies the work involved in finding derivatives, especially when variables are raised to significant powers. This reliability and ease of use make it a valuable tool for calculus students.
Other exercises in this chapter
Problem 46
Find \(f^{\prime}(x)\) for the given function. \(f(x)=\sqrt{2 x+1}\) (See Exercise 55 in Section l.3.)
View solution Problem 46
Given $$ s(t)=t^{3}+t $$ where \(s(t)\) is in feet and \(t\) is in seconds, find each of the following. a) \(v(t)\) b) \(a(t)\) c) The velocity and acceleration
View solution Problem 46
Differentiate each function. \(f(x)=\frac{3 x^{2}+2 x}{x^{2}+1}\)
View solution Problem 47
Find the simplified difference quotient for each function listed. $$ f(x)=a x^{2}+b x+c $$
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