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Chapter 3: Problem 4

How do you find the derivative of the sum of two functions \(f+g ?\)

Short Answer

Expert verified
Answer: The derivative of the sum of two functions, f and g, can be found using the sum rule of differentiation, which states that the derivative of the sum is equal to the sum of their derivatives. Mathematically, this is represented as (f+g)'(x) = f'(x) + g'(x).

Step by step solution

01

Identify the functions

We are given two functions, \(f\) and \(g\). We need to find the derivative of the sum of these functions, denoted as \((f + g)'\).
02

Apply the sum rule of differentiation

According to the sum rule of differentiation, the derivative of the sum of two functions is equal to the sum of their derivatives. Mathematically, this can be written as: \((f+g)' = f' + g'\)
03

Find the derivatives of f and g

In this step, differentiate both functions \(f\) and \(g\) with respect to the variable they depend on, usually denoted as \(x\). Let \(f'(x)\) be the derivative of function \(f\) with respect to \(x\), and \(g'(x)\) be the derivative of function \(g\) with respect to \(x\).
04

Combine the derivatives of f and g

Now, add the derivatives obtained in step 3 to find the derivative of the sum of the two functions \(f+g\): \((f+g)'(x) = f'(x) + g'(x)\) This is the final answer for the derivative of the sum of two functions \(f+g\).

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Key Concepts

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

Sum Rule of Differentiation
The sum rule of differentiation is a fundamental principle in calculus that simplifies finding the derivative of the sum of two or more functions. This rule states that the derivative of a sum is equal to the sum of the derivatives.

Mathematically, if you have two functions, say \( f(x) \) and \( g(x) \), the sum rule can be expressed as:
  • \((f + g)' = f' + g'\)
This means, to find the derivative of \( f + g \), you simply find \( f' \) and \( g' \) separately and then add them together. This rule makes it very easy to handle more complex expressions by breaking them down into simpler parts.

Whenever tackling a differentiation problem where functions are added together, remembering the sum rule can save you time and effort by clearly indicating that differentiation can be performed term by term.
Differentiation
Differentiation is one of the core operations in calculus. It involves finding the derivative of a function, which represents the rate of change of the function with respect to a variable, typically \( x \).

By differentials, we mean how a small change in one variable results in a change in another variable. This concept aids in comprehending how functions vary and in realizing the slope of the function at any particular point on its curve.

Key points to remember about differentiation:
  • It's used to find rates of change, slopes of curves, and model varying processes.
  • Chain rule, product rule, and quotient rule are other differentiation rules often used alongside the sum rule.
  • Differentiation helps in optimizing problems, finding maximums and minimums of functions.
Understanding the basics of differentiation equips you to handle a wide range of mathematical problems efficiently and is crucial for success in calculus.
Calculus
Calculus is an essential field of mathematics that deals with change and motion; it is divided mainly into two branches: differentiation and integration.

Differentiation focuses on the concept of derivatives, helping us understand how things change at an instant, while integration deals with the sum of small units that lead to an accumulation, like areas under curves.

Benefits of learning calculus:
  • It allows you to model and solve real-world problems in physics, engineering, economics, and more.
  • Calculus has applications in diverse fields like computer science, statistics, biology, and chemistry.
  • It enhances analytical and problem-solving skills, preparing you for advanced studies in mathematics.
Mastering calculus involves understanding its core concepts, such as limits, functions, derivatives, integrals, and infinite series, all of which are interconnected and essential for thorough comprehension.

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Most popular questions from this chapter

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u \end{array}\right.$$ a. Show that \(\lim _{y \rightarrow u} H(v)=0\) b. For any value of \(u,\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)$$ c. Show that $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right)$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\)

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$6 x^{3}+7 y^{3}=13 x y$$

Suppose \(f\) is differentiable for all real numbers with \(f(0)=-3, f(1)=3, f^{\prime}(0)=3,\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=\sin (\pi f(x)) .\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}(1)\)

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(y^{2}(x+2)=x^{2}(6-x)\) (trisectrix)

A study conducted at the University of New Mexico found that the mass \(m(t)\) (in grams) of a juvenile desert tortoise \(t\) days after a switch to a particular diet is described by the function \(m(t)=m_{0} e^{0.004 t},\) where \(m_{0}\) is the mass of the tortoise at the time of the diet switch. If \(m_{0}=64\) evaluate \(m^{\prime}(65)\) and interpret the meaning of this result. (Source: Physiological and Biochemical Zoology, 85,1,2012 )
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