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Suppose that the function \(v\) in the Derivative Product Rule has a constant value \(c .\) What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?

Short Answer

Expert verified
The Product Rule simplifies to the Constant Multiple Rule when one function is constant.

Step by step solution

01

Recall the Derivative Product Rule

The Derivative Product Rule states: If you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by \( (uv)' = u'v + uv' \). This rule is used to differentiate the product of two functions.
02

Substitute the Constant Function

Since \(v(x) = c\) is a constant, its derivative \(v'(x) = 0\). Substitute \(v(x)\) and \(v'(x)\) into the Product Rule formula: \( (uc)' = u'c + uc' \).
03

Simplify the Expression

Given \(c' = 0\), the term \(uc' = u\cdot0 = 0\). This simplifies the expression to \((uc)' = u'c + 0 = u'c\). Hence, the derivative simplifies to the constant multiplied by the derivative of \(u(x)\).
04

Understand the Derivative Constant Multiple Rule

The Derivative Constant Multiple Rule states that if \(c\) is a constant and \(u(x)\) is a function, then the derivative \((cu)' = c \, u'(x)\). The simplification in the previous step matches this rule exactly, helping prove the consistency of the derivative rules.
05

Conclude the Relationship

When \(v(x)\) in the Product Rule is a constant, the rule effectively becomes the Constant Multiple Rule, thereby confirming there's no contradiction between the rules. This highlights that multiplying a function by a constant simply scales its derivative by that constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Constant Multiple Rule
The Derivative Constant Multiple Rule is a fundamental concept in calculus that simplifies finding the derivative of a product involving a constant. Essentially, this rule tells us that if you multiply a function by a constant, the derivative of the resulting function is the constant multiplied by the derivative of the original function. This rule is represented mathematically as \((cu)' = c \cdot u'(x)\).

This means:
  • "\(c\)" stays as it is - you don’t differentiate the constant.
  • "\(u(x)\)" is differentiated normally - find its derivative \(u'(x)\).
  • Combine them by multiplying the constant \(c\) with \(u'(x)\).
This rule is particularly useful because it simplifies calculations for functions where part of the expression is constant. It helps in understanding that constants don't affect the rate of change of a function—they only affect the magnitude of this change.

This concept reinforces how constants and variables interact in differentiation, showing us that multiplying a function by a constant amplifies its rate of change by that constant amount. Hence, the core idea is about scaling functions while maintaining the nature of the derivative.
differentiation
Differentiation is one of the core concepts in calculus and represents the process of finding the derivative of a function. The derivative is a measure of how a function changes as its input changes. It's essentially the rate at which one quantity changes with respect to another.

Let's break down the basics:
  • The derivative of a function \(f(x)\) is denoted as \(f'(x)\) or \( \frac{df}{dx} \).
  • It tells you how steep the function is at any given point, or the slope of the tangent line at that point.
  • It can indicate increasing or decreasing trends and even the acceleration or deceleration.
This concept is used widely across different fields such as physics, engineering, economics, and more, because it helps to predict future behavior of quantities, analyze trends, and solve optimization problems.

In mathematical terms, differentiation involves applying rules like the Constant Multiple Rule, Power Rule, Product Rule, among others, to calculate these changes efficiently. By mastering differentiation, we gain deep insights into the behavior of functions and their applications.
calculus concepts
Calculus is a branch of mathematics focused on the study of change. At its core are two fundamental operations: differentiation and integration. Both have profound implications in mathematics and its applications.

Here are key aspects of calculus concepts relevant to differentiation:
  • Differentiation: Focuses on finding the derivative, which shows the rate of change or slope at any point of a function.
  • Integration: The reverse process of differentiation, used to find the area under the curves and accumulate quantities.
  • Rules and Theorems: Various rules such as Derivative Rules (Constant Multiple, Product, etc.) help in systematically differentiating complex equations.
Calculus uses these operations to model and solve problems in real-world scenarios;
from simple physics problems of motion to complex financial models predicting stock trends.

Understanding calculus concepts allows you to transition from calculations and formulas to application and analysis, making it a powerful tool in both pure and applied sciences.

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Most popular questions from this chapter

If \(x^{2}+y^{2}=25\) and \(d x / d t=-2,\) then what is \(d y / d t\) when \(x=3\) and \(y=-4 ?\)

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A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. \begin{equation} x y^{3}+x^{2} y=6 \end{equation}

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 \(k T / 2 .\)

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