Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 37E
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Question
![**Problem Statement:**
**Find the equation of the tangent line to the curve \( y = x^4 + 2e^x \) at the point \( (0, 2) \). (Hint: Don’t forget that the derivative of \(e^x\) is \(e^x\).)**
**Answer Box:**
\[ y = \] [_______________]
**Explanation:**
To find the equation of the tangent line to the curve at the given point, follow these steps:
1. **Find the derivative of the function \( y = x^4 + 2e^x \):**
The derivative, \( y' \), represents the slope of the tangent line at any point \( x \) on the curve.
\[
y = x^4 + 2e^x \implies y' = \frac{d}{dx}(x^4) + \frac{d}{dx}(2e^x)
\]
Using the power rule and the exponential rule for differentiation:
\[
\frac{d}{dx}(x^4) = 4x^3 \quad \text{and} \quad \frac{d}{dx}(2e^x) = 2e^x
\]
Hence, the derivative is:
\[
y' = 4x^3 + 2e^x
\]
2. **Evaluate the derivative at the given point \( (0, 2) \):**
Substitute \( x = 0 \) into \( y' \):
\[
y'(0) = 4(0)^3 + 2e^0 = 0 + 2 \times 1 = 2
\]
Therefore, the slope \( m \) of the tangent line at the point \( (0, 2) \) is 2.
3. **Use the point-slope form of the equation of a line:**
The point-slope form is given by:
\[
y - y_1 = m(x - x_1)
\]
Here, \( (x_1, y_1) = (0, 2) \) and \( m = 2 \):
\[
y - 2 = 2(x - 0) \implies y - 2 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F114d112a-89da-40ea-8ec8-a25f26317aff%2Fda0aef7e-2d94-494b-9073-347f6b5dfdce%2Fzni6dx_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
**Find the equation of the tangent line to the curve \( y = x^4 + 2e^x \) at the point \( (0, 2) \). (Hint: Don’t forget that the derivative of \(e^x\) is \(e^x\).)**
**Answer Box:**
\[ y = \] [_______________]
**Explanation:**
To find the equation of the tangent line to the curve at the given point, follow these steps:
1. **Find the derivative of the function \( y = x^4 + 2e^x \):**
The derivative, \( y' \), represents the slope of the tangent line at any point \( x \) on the curve.
\[
y = x^4 + 2e^x \implies y' = \frac{d}{dx}(x^4) + \frac{d}{dx}(2e^x)
\]
Using the power rule and the exponential rule for differentiation:
\[
\frac{d}{dx}(x^4) = 4x^3 \quad \text{and} \quad \frac{d}{dx}(2e^x) = 2e^x
\]
Hence, the derivative is:
\[
y' = 4x^3 + 2e^x
\]
2. **Evaluate the derivative at the given point \( (0, 2) \):**
Substitute \( x = 0 \) into \( y' \):
\[
y'(0) = 4(0)^3 + 2e^0 = 0 + 2 \times 1 = 2
\]
Therefore, the slope \( m \) of the tangent line at the point \( (0, 2) \) is 2.
3. **Use the point-slope form of the equation of a line:**
The point-slope form is given by:
\[
y - y_1 = m(x - x_1)
\]
Here, \( (x_1, y_1) = (0, 2) \) and \( m = 2 \):
\[
y - 2 = 2(x - 0) \implies y - 2 =
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