Problem 8
Question
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=e^{2 x / 3}\end{equation}
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dy}{dx} = \frac{2}{3} e^{\frac{2x}{3}} \).
1Step 1: Identify the Function
The function given is \[ y = e^{\frac{2x}{3}} \]This is an exponential function where the exponent is a linear function of \(x\). We need to find its derivative with respect to \(x\).
2Step 2: Apply the Chain Rule
To differentiate \( y = e^{\frac{2x}{3}} \), use the chain rule. The chain rule states that\[ \frac{dy}{dx} = \frac{d}{dx} \left( e^{u} \right) \cdot \frac{du}{dx}, \]where \( u = \frac{2x}{3} \).
3Step 3: Differentiate the Exponential Function
The derivative of \( e^{u} \) with respect to \( u \) is itself, \( e^{u} \). Therefore, \[ \frac{d}{du} \left( e^{\frac{2x}{3}} \right) = e^{\frac{2x}{3}}. \]
4Step 4: Differentiate the Inner Function
Now, find the derivative of \( u = \frac{2x}{3} \) with respect to \( x \). The derivative is a constant multiplied by a variable, so \[ \frac{du}{dx} = \frac{2}{3}. \]
5Step 5: Combine the Results
Combining the results from step 3 and step 4 using the chain rule gives:\[ \frac{dy}{dx} = e^{\frac{2x}{3}} \cdot \frac{2}{3}. \] Thus, \[ \frac{dy}{dx} = \frac{2}{3} e^{\frac{2x}{3}}. \]
Key Concepts
Exponential FunctionChain RuleCalculusDifferentiation
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In our exercise, the function is given by \( y = e^{\frac{2x}{3}} \). Here, \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
In exponential functions:
In exponential functions:
- The base \( e \) remains constant.
- The variable (in this case \( x \)) is in the exponent position.
- Exponential functions are significant in many real-world applications such as compound interest, population growth, and more.
- The rate of change (or growth rate) is proportional to the value of the function itself.
- The derivative of \( e^u \) with respect to \( u \) is itself, \( e^u \).
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions, which are functions composed of other functions. Suppose you have a function \( y \) that includes another function \( u(x) \) inside it. The chain rule helps find the derivative of such functions.
A typical scenario is when you have \( y = f(g(x)) \). The chain rule states:
A typical scenario is when you have \( y = f(g(x)) \). The chain rule states:
- Differentiate the outer function \( f \) with respect to the inner function \( u \).
- Multiply this derivative by the derivative of the inner function \( u \) with respect to the variable, \( x \).
- \( u = \frac{2x}{3} \), an inner linear function.
- The derivative of \( u \) is \( \frac{2}{3} \).
Calculus
Calculus is the branch of mathematics that explores concepts like change and motion, primarily through the study of derivatives and integrals. It provides a framework for modeling systems with changing quantities.
Key areas in calculus include:
Key areas in calculus include:
- Differentiation: Focuses on how functions change, finding rates of change, and the slope of curves.
- Integration: Deals with finding areas under curves and accumulated quantities.
Differentiation
Differentiation is a process in calculus used to find the derivative of a function. The derivative measures how a function's output value changes with changes in its input value. It calculates the slope of the tangent line to the curve of the function at any given point.
Here are a few key points about differentiation:
Here are a few key points about differentiation:
- The derivative tells us the rate of change of the function.
- It can determine if the function is increasing, decreasing, or constant over a region.
- In practical terms, it's used in physics for velocity, in economics for cost functions, and more.
Other exercises in this chapter
Problem 8
In Exercises \(5-8,\) show that each function is a solution of the given initial value problem. $$\begin{array}{ll}{x^{2} y^{\prime}=x y-y^{2},} & {y(e)=e \quad
View solution Problem 8
Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{x \rightarrow-5} \frac{x^{2}-25}{x+5} $$
View solution Problem 9
Find the values in Exercises \(9-12\) $$ \sin \left(\cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)\right) $$
View solution Problem 9
True, or false? As \(x \rightarrow \infty\) $$ \begin{array}{ll}{\text { a. }} & {x=o(x)} & {\text { b. } x=o(x+5)} \\\ {\text { c. }} & {x=O(x+5)} & {\text { d
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