Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\arccos \left(e^{x}\right) $$

Step-by-Step Solution

Answer

\( D_x = -\frac{e^x}{\sqrt{1-e^{2x}}} \)

Step 1: Identify the Derivative Rule

The function we have is a composite function where the outer function is the arccosine and the inner function is the exponential function. For finding the derivative, we need to use the chain rule.

Step 2: Differentiate the Outer Function

The derivative of the outer function \(\arccos(u)\) with respect to \(u\) is \(-\frac{1}{\sqrt{1-u^2}}\). Here, \(u = e^x\).

Step 3: Differentiate the Inner Function

The derivative of the inner function \(e^x\) with respect to \(x\) is \(e^x\).

Step 4: Apply the Chain Rule

Using the chain rule, the derivative of \(y = \arccos(e^x)\) with respect to \(x\) is \(-\frac{1}{\sqrt{1-(e^x)^2}}\cdot e^x\).

Step 5: Simplify the Expression

Substituting \(u = e^x\) into our expression gives us \(-\frac{e^x}{\sqrt{1-(e^{2x})}}\). So, \(D_x y = -\frac{e^x}{\sqrt{1-e^{2x}}}\).

Key Concepts

Chain Rule

The chain rule is a fundamental rule in calculus that is essential when dealing with composite functions. Simply put, the chain rule helps us differentiate a function that contains another function inside it. This is a bit like peeling an onion, where you need to find the derivative of the outer layer first, and then work your way inside.
In our exercise, we have a composite function: \(y = \arccos(e^x)\). This function involves the composition of the arccosine function and the exponential function.
Here’s how the chain rule works:

• Differentiating the outer function: We first find the derivative of the arccosine function. When differentiating \(\arccos(u)\) with respect to \(u\), we get \(-\frac{1}{\sqrt{1-u^2}}\).
• Differentiating the inner function: Next, we find the derivative of the inner function \(e^x\) with respect to \(x\), which is simply \(e^x\).
• Combining these using the chain rule: Finally, applying the chain rule means multiplying these two derivatives together: \(-\frac{1}{\sqrt{1-(e^x)^2}}\cdot e^x\).

This method allows us to handle complex functions easily and find their derivatives with precision.

Composite Functions

A composite function is basically a function within a function. It’s like a nesting doll—once you open one, there’s another one inside. When we write a composite function, we usually denote it as \(f(g(x))\) where \(g(x)\) is the inner function and \(f\) is the outer function.
In this exercise, the composite function is \(y = \arccos(e^x)\). Here, \(e^x\) is nestled inside the arccosine function, or in other words, \(e^x\) is the inner function and \(\arccos{(\cdot)}\) is the outer function.
Understanding composite functions is crucial because they help in modeling problems where two variables are connected indirectly, and they appear frequently in calculus. To solve them, we often rely on the chain rule to differentiate them, as each layer of function affects the overall change in \(y\) with respect to \(x\).

Derivative of Exponential Functions

Exponential functions have a unique property where their derivatives look remarkably similar to the original function. This is because the derivative of \(e^x\) is \(e^x\) itself. This characteristic makes them simpler to work with compared to other functions.
In our problem, the inner function \(e^x\) is an exponential function. So, when differentiating this part with respect to \(x\), we simply get \(e^x\). This elegant property of the exponential function contributes to the overall result when we apply the chain rule.
This derivative property is extremely handy in calculus, particularly when dealing with differential equations and growth models, where exponential functions often describe rapid growth and decay processes. Remembering that the derivative of \(e^x\) is itself can save a lot of time and reduce complexity in solving problems involving exponential functions.