Chapter 3: Problem 11
find the indicated derivatives. $$\frac{d p}{d q} \text { if } p=q^{3 / 2}$$
Short Answer
Expert verified
The derivative \( \frac{dp}{dq} = \frac{3}{2}q^{1/2} \).
Step by step solution
01
Understanding the Problem
We're asked to find the derivative of a function defined as \( p = q^{3/2} \) with respect to \( q \). This means we need \( \frac{dp}{dq} \).
02
Applying the Power Rule
The function \( p = q^{3/2} \) is a power function. We can apply the power rule for derivatives, which states that if \( y = x^n \), then \( \frac{dy}{dx} = n \cdot x^{n-1} \).
03
Calculating the Derivative
Using the power rule, for \( p = q^{3/2} \), the derivative will be \( \frac{dp}{dq} = \frac{3}{2} \cdot q^{(3/2)-1} \).
04
Simplifying the Expression
Simplify \( \frac{3}{2} \cdot q^{1/2} \) to get the final expression of the derivative. So, \( \frac{dp}{dq} = \frac{3}{2} \cdot q^{1/2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Power Rule
The power rule is a fundamental shortcut in calculus for finding derivatives of functions that have a variable raised to a power. It's very handy and simplifies things greatly. The rule states that if you have a function in the form of \( y = x^n \), where \( n \) is any real number, the derivative is \( \frac{dy}{dx} = n \cdot x^{n-1} \).
It's like saying: *"Take the exponent, bring it down as a multiplier, and then reduce that exponent by one."*
This rule is particularly useful because it works on a wide range of functions, from simple polynomials to more complex expressions involving roots and fractions. Understanding and applying the power rule saves a lot of time compared to calculating derivatives from first principles.
It's like saying: *"Take the exponent, bring it down as a multiplier, and then reduce that exponent by one."*
This rule is particularly useful because it works on a wide range of functions, from simple polynomials to more complex expressions involving roots and fractions. Understanding and applying the power rule saves a lot of time compared to calculating derivatives from first principles.
- The power rule is not just for positive integers; it applies to negative exponents and fractions, as seen in our exercise.
- Don't worry about messy fractions or square roots, the power rule can handle them.
Calculus and Its Importance
Calculus is a branch of mathematics that helps us understand changes between values that are related by a function. It consists of two main parts: differentiation and integration. Differentiation is about finding the rate of change, while integration accumulates quantities.
In our exercise, we're focusing on differentiation, a key tool in calculus that lets us find rates of change. It's not just used in math class; it has applications ranging from science and engineering to economics and beyond!
In our exercise, we're focusing on differentiation, a key tool in calculus that lets us find rates of change. It's not just used in math class; it has applications ranging from science and engineering to economics and beyond!
- Calculus helps describe phenomena in the physical world, such as motion, heat, light, harmonics, and electricity.
- It provides the tools for modeling and solving equations that describe changing scenarios, like weather patterns and financial models.
Basics of Differentiation
Differentiation is the process of finding a derivative, which essentially measures how a function changes as its input changes. This is vital for understanding the behavior of mathematical models.
When we differentiate a function, we produce a new function that gives the slope of the original function at any given point. This can indicate where the function increases or decreases and by how much.
When we differentiate a function, we produce a new function that gives the slope of the original function at any given point. This can indicate where the function increases or decreases and by how much.
- In simple terms, the derivative tells us the rate at which one quantity changes with respect to another.
- For the function \( p = q^{3/2} \), differentiation helped us find \( \frac{dp}{dq} = \frac{3}{2} \cdot q^{1/2} \), showing the rate of change of \( p \) with respect to \( q \).