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Identify an inner function \(u=g(x)\) and an outer function \(y=f(u)\) of \(y=e^{x^{3}+2 x} .\) Then calculate \(\frac{d y}{d x}\) using \(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}.\)

Short Answer

Expert verified
Question: Find the derivative of the function \(y = e^{x^3 + 2x}\) using the chain rule. Answer: The derivative of the function \(y = e^{x^3 + 2x}\) is \(\frac{dy}{dx} = e^{x^3 + 2x} \cdot (3x^2 + 2)\).

Step by step solution

01

Identify the inner and outer functions

Consider the given function \(y = e^{x^3 + 2x}\), we can identify the inner function \(u = g(x)\) as the exponent of \(e\), which is \(u = g(x) = x^3 + 2x\). The outer function \(y = f(u)\) is the exponential function with base \(e\), so \(y = f(u) = e^u\).
02

Calculate \(\frac{dy}{du}\)

Now that we know the outer function \(y = f(u) = e^u\), we can find its derivative with respect to \(u\). Notice that the derivative of \(e^u\) with respect to \(u\) is just itself. So \(\frac{dy}{du} = e^u\).
03

Calculate \(\frac{du}{dx}\)

Next, we need to find the derivative of the inner function \(u = x^3 + 2x\) with respect to \(x\). Applying the power rule and sum rule, we get \(\frac{du}{dx} = 3x^2 + 2\).
04

Use the chain rule to find \(\frac{dy}{dx}\)

Finally, we'll use the chain rule formula, which states that \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). From Steps 2 and 3, we know that \(\frac{dy}{du} = e^u\) and \(\frac{du}{dx} = 3x^2 + 2\). Therefore, we have: \[ \frac{dy}{dx} = e^u \cdot (3x^2 + 2) = e^{x^3 + 2x} \cdot (3x^2 + 2) \] So, the derivative \(\frac{dy}{dx}\) of the given function \(y = e^{x^3 + 2x}\) is given by: \[ \frac{dy}{dx} = e^{x^3 + 2x} \cdot (3x^2 + 2) \]

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Key Concepts

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

Derivative
Understanding derivatives is fundamental in calculus. A derivative represents the rate of change of a function concerning its variable. Think of it as the function's "instantaneous speed." When you calculate a derivative, you're finding out how fast or slow something is changing at any given point.There are several rules and methods to find derivatives efficiently:
  • Power Rule: This is used when differentiating expressions like \( x^n \). The derivative becomes \( nx^{n-1} \).
  • Sum Rule: This helps find the derivative of a sum of functions. It tells us that the derivative of \( f(x) + g(x) \) is just the derivative of \( f(x) \) plus that of \( g(x) \).
  • Product and Quotient Rules: These are for finding derivatives of products or quotients of functions, which combine the derivatives of individual functions in a specific way.
In essence, the derivative gives a snapshot of the behavior of a function, helping us understand trends and changes. Mastery of derivatives opens up a deeper understanding of the world described by mathematics.
Composite Functions
Composite functions are formed by putting one function inside another. An easy way to visualize this is imagining a machine within another machine. You input something into the first, and its output feeds directly into the second.Mathematically, if \( y = f(g(x)) \), \( g(x) \) is the inner function and \( f(u) \) is the outer function where \( u=g(x) \). We evaluate \( f \) at the result of \( g(x) \). This nesting can be deeper, but understanding simple two-layer compositions is crucial.The chain rule, which is vital when dealing with derivative of composite functions, essentially states:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]This means that to take the derivative of a composite function, you take the derivative of the outer function with respect to the inner function, and multiply it by the derivative of the inner function with respect to its variable.By mastering composite functions, you'll gain the ability to handle more complex mathematical expressions and applications.
Exponential Function
Exponential functions are a special type of function where a constant base is raised to a variable exponent. They appear in various fields like finance, biology, and physics.For example, the function \( e^x \) is the most common exponential function, where \( e \) (approximately 2.718) is the base of the natural logarithm. It is unique because if we take its derivative, it remains unchanged: \[\frac{d}{dx}(e^x) = e^x\]This property makes the exponential function incredibly powerful, especially in calculus and differential equations.When an exponential function has a more complex exponent, like \( e^{x^3 + 2x} \), it becomes a composite function. Here, the exponential part acts on an additional function within, requiring the application of the chain rule to find its derivative.Understanding exponential functions allows you to solve complex problems involving rapid growth or decay processes in various scientific and engineering contexts.

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Most popular questions from this chapter

Graph the functions \(f(x)=x^{3}, g(x)=3^{x}\) End \(h(x)=x^{x}\) and find their common intersection point (exactly).

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

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Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. Find the second derivative \(\frac{d^{2} y}{d t^{2}}\) b. Verify that \(\frac{d^{2} y}{d t^{2}}=-\frac{k}{m} y\)

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