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What is the derivative of \(y=e^{k x} ?\) For what values of \(k\) does this rule apply?

Short Answer

Expert verified
Answer: The derivative of the function \(y = e^{kx}\) with respect to \(x\) is \(\frac{dy}{dx} = ke^{kx}\). This rule applies for any real value of \(k\).

Step by step solution

01

Recall the chain rule for differentiation

The chain rule for differentiation states that if a function \(y\) can be represented as a composition of two functions \(u\) and \(v\), i.e., \(y=u(v(x))\), then the derivative of \(y\) with respect to \(x\) is given by: $$\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$$
02

Identify the composition of functions

In this case, we can represent the given function as a composition of two functions by assigning \(u=e^v\) and \(v=kx\). This means that \(y=u(v(x))=e^{kx}\).
03

Find the derivative of each function

Now, we will find the derivatives of each function with respect to their respective variables: For \(u=e^v\), the derivative is: $$\frac{du}{dv} = e^v$$ For \(v=kx\), the derivative is: $$\frac{dv}{dx} = k$$
04

Apply the chain rule

Using the chain rule for the given function \(y=e^{kx}\), its derivative is: $$\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = e^v \cdot k = e^{kx} \cdot k$$ So, the derivative of \(y=e^{kx}\) is: $$\frac{dy}{dx} = ke^{kx}$$
05

Discuss the values of \(k\) for which the rule applies

The chain rule for differentiation applies to any real value of \(k\). The derivative obtained, \(ke^{kx}\), is valid for any real value of \(k\).

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Key Concepts

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

Derivative
The concept of a derivative is central to calculus and describes how a function changes as its input changes. When we talk about the derivative of a function, we're looking at the rate of change or the slope of the function at any given point. In mathematical terms, the derivative of a function \( f(x) \) is denoted by \( f'(x) \) or \( \frac{df}{dx} \). It tells us how much \( y \) (the output) changes with a small change in \( x \) (the input). For exponential functions like \( y = e^{kx} \), the derivative measures the change in \( y \) as \( x \) changes. Understanding derivatives helps in many fields, including physics for understanding motion, in economics for understanding how different factors affect growth, and in data science for optimizing algorithms.
Exponential Functions
Exponential functions are a fundamental class of functions characterized by the independent variable appearing in the exponent. A classic example of an exponential function is \( y = e^{kx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. This base gives rise to natural exponential functions which have useful properties in calculus and real-world applications.
  • Rapid Growth or Decay: The function \( y = e^{kx} \) grows or decays exponentially depending on the sign of \( k \).
  • Applications: Exponential functions model population growth, radioactive decay, interest calculations, and other processes where change occurs at a continuous rate.
When dealing with exponential functions, it's vital to recognize their unique growth patterns and how the parameter \( k \) influences their behavior, further shedding light on how specialization or focus on certain types of problems can positively impact understanding.
Differentiation
Differentiation is the process of finding a derivative, meaning determining the rate at which something changes. This process is core to calculus and is applied in diverse fields ranging from engineering to medicine.To differentiate a function like \( y = e^{kx} \), we employ rules such as the chain rule. The chain rule helps us tackle more complex functions by breaking them down into simpler parts. In this context, differentiation involves:
  • Applying the Chain Rule: We identify the composite nature of \( y = e^{kx} \) with \( u = e^v \) and \( v = kx \), and apply the rule \( \frac{du}{dx} = \frac{du}{dv} \times \frac{dv}{dx} \).
  • Executing Derivative Calculations: Calculating \( \frac{du}{dv} = e^v \) and \( \frac{dv}{dx} = k \), then combining via the chain rule to get \( \frac{dy}{dx} = ke^{kx} \).
Differentiation not only gives us the formula we seek but also strengthens our overall understanding of how functions behave and change, preparing us for more complex analysis and problem-solving in future topics.

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Most popular questions from this chapter

A lighthouse stands 500 m off a straight shore and the focused beam of its light revolves four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

Recall that \(f\) is even if \(f(-x)=f(x),\) for all \(x\) in the domain of \(f,\) and \(f\) is odd if \(f(-x)=-f(x),\) for all \(x\) in the domain of \(f\). a. If \(f\) is a differentiable, even function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither. b. If \(f\) is a differentiable, odd function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither.

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

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