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Chapter 4: Problem 6

Use the product rule to find the derivative with respect to the independent variable. \(f(x)=2\left(3 x^{2}-2 x^{3}\right)\left(1-5 x^{2}\right)\)

Short Answer

Expert verified
The derivative is \( f'(x) = 12x - 12x^2 - 120x^3 + 100x^4 \).

Step by step solution

01

Understand the problem

We are given the function \( f(x) = 2(3x^2 - 2x^3)(1 - 5x^2) \). Our goal is to find the derivative \( f'(x) \) with respect to \( x \) using the product rule.
02

Identify the components for the product rule

The product rule states that \((uv)' = u'v + uv'\). Here, let \( u = 2(3x^2 - 2x^3) \) and \( v = (1 - 5x^2) \). We need to find \( u', v' \) and then apply the product rule.
03

Find the derivative of u

Calculate \( u' \). First simplify \( u = 6x^2 - 4x^3 \). Then, the derivative \( u' = \frac{d}{dx}(6x^2 - 4x^3) = 12x - 12x^2 \).
04

Find the derivative of v

Calculate \( v' \). The derivative \( v' = \frac{d}{dx}(1 - 5x^2) = -10x \).
05

Apply the product rule

Using the product rule, \( f'(x) = u'v + uv' \). Substitute \( u', v', u, \) and \( v \) into the formula: \( f'(x) = (12x - 12x^2)(1 - 5x^2) + (6x^2 - 4x^3)(-10x) \).
06

Simplify the expression

Expand and simplify the terms: 1. \((12x - 12x^2)(1 - 5x^2) = 12x - 60x^3 - 12x^2 + 60x^4\). 2. \((6x^2 - 4x^3)(-10x) = -60x^3 + 40x^4\). Combine the expressions: \( f'(x) = 12x - 12x^2 - 120x^3 + 100x^4 \).

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Key Concepts

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

Product Rule
The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is given by: \[ (uv)' = u'v + uv' \] This means you differentiate \( u \) with respect to \( x \), multiply it by \( v \), and add it to \( u \) multiplied by the derivative of \( v \).
  • Example: Consider the functions \( u = 6x^2 - 4x^3 \) and \( v = 1 - 5x^2 \).
  • The product rule helps compute the derivative of \( f(x) = u(x)v(x) \).
Understanding how to apply the product rule is crucial for tackling complex expressions where multiple functions are multiplied. It allows you to break down the differentiation process into simpler steps, making it an indispensable tool in calculus.
Derivative
A derivative essentially measures how a function changes as its input changes. It's central to calculus and helps us understand rates of change. The notation \( f'(x) \) or \( \frac{df}{dx} \) is commonly used for derivatives.
  • Basic Idea: The derivative evaluates how one quantity changes with respect to another. For example, in the function \( y = f(x) \), the derivative tells us how \( y \) changes as \( x \) changes.
  • Computing: To find the derivative of a function like \( u = 6x^2 - 4x^3 \), use power rule techniques: bring the exponent down and subtract one. - For \( 6x^2 \), the derivative is \( 12x \). - For \( -4x^3 \), the derivative is \( -12x^2 \).
Derivatives have numerous applications in real life, such as in physics for measuring velocity and acceleration, economics for cost and revenue predictions, and many other areas.
Independent Variable
In calculus, the independent variable is the variable you change to observe how it affects another quantity. For example, in the function \( f(x) \), \( x \) is the independent variable. It is the input to the function, determining the output \( f(x) \).
  • Role: Understanding the role of the independent variable helps in interpreting graphs and results.
  • Context: It gives context to your calculations, ensuring clarity in how you apply calculus concepts like derivatives.
  • In practical terms, the independent variable is what you "control" or manipulate in experiments or equations.
The independent variable is often represented on the x-axis of a graph, highlighting its role as the foundational input to determine the behavior of the dependent variable, typically shown on the y-axis.

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Most popular questions from this chapter

The following limit represents the derivative of a function \(f\) at the point \((a, f(a))\) : $$\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{6}+h\right)-\sin \frac{\pi}{6}}{h}$$ Find \(f\) and \(a\).

(a) Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. \(\tan (0.01)\)

Suppose that the per capita growth rate of a population is \(2 \% ;\) that is, if \(N(t)\) denotes the population size at time \(t\), then $$\frac{1}{N} \frac{d N}{d t}=0.02$$ Suppose also that the population size at time \(t=2\) is equal to \(100 .\) Use a linear approximation to compute the population size at time \(t=2.1\).

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\frac{1}{1+x}\) at \(a=0\)
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