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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![Title: Calculating the Derivative of a Function
Find the derivative of the function \( f(y) \), below. It may be to your advantage to simplify first.
\[ f(y) = 4^y (12 - y^5) \]
\[ f'(y) = \] [Answer box for input]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F461e6b03-e26d-48fc-aa6c-204c3b542fc6%2F5a0a662c-ecbd-4540-8c16-c917d4a0bd88%2Fzqejh34_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Title: Calculating the Derivative of a Function
Find the derivative of the function \( f(y) \), below. It may be to your advantage to simplify first.
\[ f(y) = 4^y (12 - y^5) \]
\[ f'(y) = \] [Answer box for input]
![### Calculus: Finding the Tangent Line to a Curve
**Problem Statement:**
Find an equation for the line tangent to the graph of
\[ f(x) = -3xe^x \]
at the point \((a, f(a))\) for \( a = 2 \).
**Solution:**
First, we need to find the derivative of \( f(x) \), which will give us the slope of the tangent line at any point \( x \). The function \( f(x) \) is defined as:
\[ f(x) = -3xe^x \]
Using the product rule for differentiation, which states that for two differentiable functions \( u(x) \) and \( v(x) \),
\[ \dfrac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x), \]
we can set \( u(x) = -3x \) and \( v(x) = e^x \). Therefore,
\[ u(x) = -3x \quad \text{implies} \quad u'(x) = -3, \]
\[ v(x) = e^x \quad \text{implies} \quad v'(x) = e^x. \]
Applying the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
\[ f'(x) = (-3)e^x + (-3x)e^x \]
\[ f'(x) = -3e^x - 3xe^x \]
\[ f'(x) = -3e^x(1 + x) \]
To find the slope at \( x = 2 \):
\[ f'(2) = -3e^2(1 + 2) \]
\[ f'(2) = -3e^2 \cdot 3 \]
\[ f'(2) = -9e^2 \]
Next, substitute \( x = 2 \) into the original function to find \( f(2) \):
\[ f(2) = -3(2)e^2 \]
\[ f(2) = -6e^2 \]
We now have the slope of the tangent line \( m = -9e^2 \) and the point \((2, -6e^2)\). Using the point-slope form of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F461e6b03-e26d-48fc-aa6c-204c3b542fc6%2F5a0a662c-ecbd-4540-8c16-c917d4a0bd88%2Fnicwnvn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus: Finding the Tangent Line to a Curve
**Problem Statement:**
Find an equation for the line tangent to the graph of
\[ f(x) = -3xe^x \]
at the point \((a, f(a))\) for \( a = 2 \).
**Solution:**
First, we need to find the derivative of \( f(x) \), which will give us the slope of the tangent line at any point \( x \). The function \( f(x) \) is defined as:
\[ f(x) = -3xe^x \]
Using the product rule for differentiation, which states that for two differentiable functions \( u(x) \) and \( v(x) \),
\[ \dfrac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x), \]
we can set \( u(x) = -3x \) and \( v(x) = e^x \). Therefore,
\[ u(x) = -3x \quad \text{implies} \quad u'(x) = -3, \]
\[ v(x) = e^x \quad \text{implies} \quad v'(x) = e^x. \]
Applying the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
\[ f'(x) = (-3)e^x + (-3x)e^x \]
\[ f'(x) = -3e^x - 3xe^x \]
\[ f'(x) = -3e^x(1 + x) \]
To find the slope at \( x = 2 \):
\[ f'(2) = -3e^2(1 + 2) \]
\[ f'(2) = -3e^2 \cdot 3 \]
\[ f'(2) = -9e^2 \]
Next, substitute \( x = 2 \) into the original function to find \( f(2) \):
\[ f(2) = -3(2)e^2 \]
\[ f(2) = -6e^2 \]
We now have the slope of the tangent line \( m = -9e^2 \) and the point \((2, -6e^2)\). Using the point-slope form of
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