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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.
  • 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.
Remember, practice using the power rule with various functions to see how versatile it is.
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!
  • 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.
Calculus is powerful because it allows us to make predictions and understand the intricate dynamics of changing systems.
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.
  • 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 \( q \).
Knowing how to differentiate functions is crucial for solving real-world problems that involve rates of change, such as velocity and acceleration.

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Most popular questions from this chapter

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$ where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L .\) In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2 .\)

The diameter of a sphere is measured as \(100 \pm 1 \mathrm{cm}\) and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.

Find $$d y / d t when x=1 if y=x^{2}+7 x-5 and d x / d t=1 / 3$$

Use a CAS to perform the following steps in Exercises \(53-60 .\) \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x^{3}-x y+y^{3}=7, \quad P(2,1) \end{equation}

Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 \(\mathrm{cm} / \mathrm{min.}\) . At what rate is the plate's area increasing when the radius is 50 \(\mathrm{cm} ?\).

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