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Find the derivative of the function below. Use lower case letters. p(y) =ay.eby dp = dy

Find the derivative of the function below. Use lower case letters. p(y) =ay.eby dp = dy

Calculus: Early Transcendentals
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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**Problem Statement:** Find the derivative of the function below. Use lower case letters. \[ p(y) = ay \cdot e^{by} \] \[ \frac{dp}{dy} = \boxed{} \] **Explanation:** The task is to calculate the derivative of the given function with respect to \( y \). The function \( p(y) \) is the product of two functions, \( ay \) and \( e^{by} \), and will require the use of the product rule for differentiation. The product rule states: if \( u(y) \) and \( v(y) \) are functions of \( y \), then the derivative of their product \( u(y) \cdot v(y) \) is given by: \[ \frac{d}{dy}[u(y) \cdot v(y)] = u(y) \cdot \frac{dv}{dy} + v(y) \cdot \frac{du}{dy} \] **Solution Outline:** 1. **Identify the Functions:** - \( u(y) = ay \) - \( v(y) = e^{by} \) 2. **Differentiate Each Function:** - \( \frac{du}{dy} = a \) (derivative of \( ay \) with respect to \( y \)) - \( \frac{dv}{dy} = be^{by} \) (derivative of \( e^{by} \) with respect to \( y \)) 3. **Apply the Product Rule:** - \( \frac{dp}{dy} = ay \cdot be^{by} + e^{by} \cdot a \) 4. **Simplify:** - \( \frac{dp}{dy} = aby \cdot e^{by} + ae^{by} \) This gives the derivative of the given function.
Transcribed Image Text:**Problem Statement:** Find the derivative of the function below. Use lower case letters. \[ p(y) = ay \cdot e^{by} \] \[ \frac{dp}{dy} = \boxed{} \] **Explanation:** The task is to calculate the derivative of the given function with respect to \( y \). The function \( p(y) \) is the product of two functions, \( ay \) and \( e^{by} \), and will require the use of the product rule for differentiation. The product rule states: if \( u(y) \) and \( v(y) \) are functions of \( y \), then the derivative of their product \( u(y) \cdot v(y) \) is given by: \[ \frac{d}{dy}[u(y) \cdot v(y)] = u(y) \cdot \frac{dv}{dy} + v(y) \cdot \frac{du}{dy} \] **Solution Outline:** 1. **Identify the Functions:** - \( u(y) = ay \) - \( v(y) = e^{by} \) 2. **Differentiate Each Function:** - \( \frac{du}{dy} = a \) (derivative of \( ay \) with respect to \( y \)) - \( \frac{dv}{dy} = be^{by} \) (derivative of \( e^{by} \) with respect to \( y \)) 3. **Apply the Product Rule:** - \( \frac{dp}{dy} = ay \cdot be^{by} + e^{by} \cdot a \) 4. **Simplify:** - \( \frac{dp}{dy} = aby \cdot e^{by} + ae^{by} \) This gives the derivative of the given function.
Expert Solution
Step 1: Given the information

The given function p open parentheses y close parentheses equals a y times e to the power of b y end exponent.

The aim is to find the derivative of the function fraction numerator d p over denominator d y end fraction.

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