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Chapter 3: Problem 28

Use Version 2 of the Chain Rule to calculate the derivatives of the following composite functions. $$y=e^{\tan t}$$

Short Answer

Expert verified
Answer: The derivative of the composite function \(y = e^{\tan t}\) with respect to \(t\) is \(\frac{dy}{dt} = e^{\tan t} \sec^2 t\).

Step by step solution

01

Identify the inner and outer functions

In the composite function \( y = e^{\tan t}\), we can see that the inner function is \(\tan t\) (which we'll call \(u\)), and the outer function is \(e^x\) where \(x = \tan t\).
02

Find the derivative of the inner function

We need to find the derivative of \(\tan t\) with respect to \(t\). The derivative of \(\tan t\) is \(\sec^2 t\). So, we have: $$\frac{du}{dt} = \sec^2 t$$
03

Find the derivative of the outer function

Now, we need to find the derivative of \(e^x\) where \(x = \tan t\). The derivative of \(e^x\) is \(e^x\). So, we have: $$\frac{de^u}{du} = e^u$$ Since \(u = \tan t\), we can replace \(u\) with \(\tan t\): $$\frac{de^u}{du} = e^{\tan t}$$
04

Use the Chain Rule to find the derivative of the composite function

Now, we can apply the Chain Rule by multiplying the derivatives in step 2 and step 3: $$\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$$ Plugging in the derivatives we found in step 2 and step 3: $$\frac{dy}{dt} = e^{\tan t} \cdot \sec^2 t$$ That gives us our final answer: $$\frac{dy}{dt} = e^{\tan t} \sec^2 t$$

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Key Concepts

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

Composite Functions
Composite functions are a way to combine two or more functions together to create a new function. These combinations involve applying one function to the results of another function. For example, in the given problem, the function is defined as \( y = e^{\tan t} \), which is a composite function where the trigonometric function \( \tan t \) is nested inside the exponential function \( e^x \).
  • The inner function, \( \tan t \), is the first function applied.
  • The outer function, \( e^x \) where \( x = \tan t \), is the second function applied to the result of the inner function.

Understanding how to separate a composite function into its constituent inner and outer functions is crucial for using the chain rule effectively in calculus.
Derivatives
In calculus, a derivative represents the rate of change of a function with respect to a variable. When working with composite functions, finding derivatives can be more challenging. This is where the chain rule becomes helpful.
  • The derivative of a function gives us information about its slope or the rate at which a function is changing at a given point.
  • For the function \( y = e^{\tan t} \), we are looking for how \( y \) changes concerning variable \( t \).

To calculate derivatives of composite functions, we make use of the chain rule to break down the problem into simpler parts, compute the derivative for each part, and then combine the results.
Inner and Outer Functions
Understanding the concepts of inner and outer functions is crucial when dealing with composite functions. By identifying these components, we can apply the chain rule to find the derivative.
  • Inner function: For our composite function \( y = e^{\tan t} \), the inner function is \( \tan t \). The change in the angle \( t \) affects \( \tan t \), and subsequently \( y \).
  • Outer function: The outer function in this context is \( e^x \), where \( x = \tan t \). This means the result of \( e^x \) depends on the value of \( \tan t \).

Knowing the inner and outer functions allows us to methodically use the chain rule to derive the derivative by first differentiating the inner function, then the outer, and finally combining these results.
Calculus
Calculus is a branch of mathematics that deals with rates of change and accumulation. Two of the main concepts in calculus are derivatives and integrals, but here, we focus on derivatives, especially in relation to composite functions.
  • One primary application of calculus is finding the slope of a curve at any given point, which is achieved through derivatives.
  • By understanding how to compute derivatives using techniques like the chain rule, calculus allows us to solve a wide array of real-world problems related to change and motion.
  • In our exercise, calculus is used to find how the function \( y = e^{\tan t} \) changes as \( t \) changes, illustrating the abstract yet practical power of calculus.

In this way, calculus provides the framework and tools needed to work with complex relationships between variables and functions.

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Most popular questions from this chapter

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

Means and tangents Suppose \(f\) is differentiable on an interval containing \(a\) and \(b,\) and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x},\) show that \(c=\sqrt{a b},\) the geometric mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi\) rad/min and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{x f(x)}{g(x)}\right]\right|_{x=4}$$
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