Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem 2: Find \( y' \) if \( y = \tan(e^{x^2}) \).**
In this problem, you are required to differentiate the function \( y = \tan(e^{x^2}) \). This involves applying the chain rule and the derivative of the tangent function.
**Solution Steps:**
1. Recognize that \( y = \tan(u) \) where \( u = e^{x^2} \).
2. Differentiate \( y = \tan(u) \) with respect to \( u \):
\[
\frac{dy}{du} = \sec^2(u)
\]
3. Differentiate \( u = e^{x^2} \) with respect to \( x \):
\[
\frac{du}{dx} = e^{x^2} \cdot 2x
\]
4. Apply the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \sec^2(e^{x^2}) \cdot e^{x^2} \cdot 2x
\]
The derivative of the function \( y = \tan(e^{x^2}) \) is thus \( y' = \sec^2(e^{x^2}) \cdot e^{x^2} \cdot 2x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f5b0569-83e9-4f3d-93a8-6d46166ffdc9%2Fed7a4a98-48dc-4c2b-8ca3-8518cfb2a944%2Fgqeouem_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Find \( y' \) if \( y = \tan(e^{x^2}) \).**
In this problem, you are required to differentiate the function \( y = \tan(e^{x^2}) \). This involves applying the chain rule and the derivative of the tangent function.
**Solution Steps:**
1. Recognize that \( y = \tan(u) \) where \( u = e^{x^2} \).
2. Differentiate \( y = \tan(u) \) with respect to \( u \):
\[
\frac{dy}{du} = \sec^2(u)
\]
3. Differentiate \( u = e^{x^2} \) with respect to \( x \):
\[
\frac{du}{dx} = e^{x^2} \cdot 2x
\]
4. Apply the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \sec^2(e^{x^2}) \cdot e^{x^2} \cdot 2x
\]
The derivative of the function \( y = \tan(e^{x^2}) \) is thus \( y' = \sec^2(e^{x^2}) \cdot e^{x^2} \cdot 2x \).
![**Problem 1:** Use implicit differentiation to find \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \).
Given equation:
\[ y^2 + xy = 2. \]
**Explanation:**
The task involves using implicit differentiation to find the derivative \(\frac{dy}{dx}\) at the specific point \((x, y) = (1, 1)\). The equation provided, \(y^2 + xy = 2\), involves both \(x\) and \(y\).
Implicit differentiation is used here because \(y\) is not expressed explicitly as a function of \(x\). Instead, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\).
Steps:
1. Differentiate each term.
2. Apply the product rule to the term \(xy\).
3. Solve for \(\frac{dy}{dx}\).
4. Substitute \(x = 1\) and \(y = 1\) to find the specific derivative value.
This process will reveal the rate of change of \(y\) with respect to \(x\) at the given point.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f5b0569-83e9-4f3d-93a8-6d46166ffdc9%2Fed7a4a98-48dc-4c2b-8ca3-8518cfb2a944%2Fc04vifu_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 1:** Use implicit differentiation to find \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \).
Given equation:
\[ y^2 + xy = 2. \]
**Explanation:**
The task involves using implicit differentiation to find the derivative \(\frac{dy}{dx}\) at the specific point \((x, y) = (1, 1)\). The equation provided, \(y^2 + xy = 2\), involves both \(x\) and \(y\).
Implicit differentiation is used here because \(y\) is not expressed explicitly as a function of \(x\). Instead, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\).
Steps:
1. Differentiate each term.
2. Apply the product rule to the term \(xy\).
3. Solve for \(\frac{dy}{dx}\).
4. Substitute \(x = 1\) and \(y = 1\) to find the specific derivative value.
This process will reveal the rate of change of \(y\) with respect to \(x\) at the given point.
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