Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=x \operatorname{arcsec}\left(x^{2}+1\right) $$

Step-by-Step Solution

Answer

\( D_x y = \text{arcsec}(x^2 + 1) + \frac{2x^2}{(x^2 + 1) \sqrt{(x^2 + 1)^2 - 1}} \)

Step 1: Write Down the Given Function

We are given the function:\[ y = x \cdot \text{arcsec}(x^2 + 1) \]Our goal is to find the derivative of \( y \) with respect to \( x \), denoted as \( D_x y \).

Step 2: Use the Product Rule

The given function is a product of two functions: \( u(x) = x \) and \( v(x) = \text{arcsec}(x^2 + 1) \). The product rule states:\[ (uv)' = u'v + uv' \]We will now differentiate each of these components.

Step 3: Differentiate the First Function

Differentiate \( u(x) = x \) with respect to \( x \):\[ u'(x) = 1 \]

Step 4: Differentiate the Second Function using the Chain Rule

Differentiate \( v(x) = \text{arcsec}(x^2 + 1) \) using the chain rule. First recall the derivative of \( \text{arcsec}(g(x)) \), which is:\[ \frac{d}{dx}[\text{arcsec}(g(x))] = \frac{g'(x)}{|g(x)|\sqrt{g(x)^2 - 1}} \]Here, \( g(x) = x^2 + 1 \), so we need \( g'(x) \).Calculate \( g'(x) \):\[ g'(x) = 2x \]Now substitute back:\[ v'(x) = \frac{2x}{(x^2 + 1)\sqrt{(x^2 + 1)^2 - 1}} \]

Step 5: Combine Results Using the Product Rule

Apply the product rule:\[ D_x y = u'v + uv' \]\[ = 1 \cdot \text{arcsec}(x^2 + 1) + x \cdot \frac{2x}{(x^2 + 1) \sqrt{(x^2 + 1)^2 - 1}} \]Simplify:\[ D_x y = \text{arcsec}(x^2 + 1) + \frac{2x^2}{(x^2 + 1) \sqrt{(x^2 + 1)^2 - 1}} \]

Key Concepts

Chain Rule

The Chain Rule is an essential tool in calculus, especially when dealing with composite functions. It's used to differentiate functions that are composed of one or more functions. In simpler terms, if you have a function nested inside another function, the Chain Rule helps to find the derivative.

To apply the Chain Rule, consider a function of the form \(y = f(g(x))\). The derivative, denoted as \(y'\), is found by differentiating the outer function \(f\) evaluated at \(g(x)\), and then multiplying it by the derivative of \(g(x)\). Mathematically, it's expressed as:

• \( y' = f'(g(x)) \cdot g'(x) \)

In our exercise, when differentiating \(v(x) = \text{arcsec}(x^2 + 1)\), the Chain Rule requires us to:

• Differentiating the outer function \(\text{arcsec}\)
• Multiplying by the derivative of the inner function \(x^2 + 1\)

This approach allows us to handle complex derivatives efficiently and is a crucial concept in calculus.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the standard trigonometric functions. They are used to find angles when the values of trigonometric functions are known. Common inverse trigonometric functions include arcsine (\(\text{arcsin}\)), arccosine (\(\text{arccos}\)), and arcsecant (\(\text{arcsec}\)).

In our problem, we work with the arcsecant function, \(\text{arcsec}(x)\). This function gives the angle whose secant is \(x\). The derivative of the arcsecant function is especially important in calculus:

• \( \frac{d}{dx}[\text{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}} \)

This derivative can look intimidating, but it’s derived by considering the relationship of arcsecant with secant and understanding how to approach these using calculus principles. In our step-by-step solution, understanding the arcsecant derivative was crucial in applying the Chain Rule correctly.

Derivative of Arcsecant

The derivative of the arcsecant function plays a pivotal role in many calculus problems, specifically those involving inverse trigonometric functions. To differentiate arcsecant, we consider a function \(y = \text{arcsec}(g(x))\), where \(g(x)\) is any function of \(x\).

The derivative is calculated as:

• \[ \frac{d}{dx}[\text{arcsec}(g(x))] = \frac{g'(x)}{|g(x)|\sqrt{g(x)^2 - 1}} \]

In this context, \(g(x) = x^2 + 1\), and its derivative \(g'(x)\) is \(2x\). Plugging these into the derivative formula gives us:

• The differentiation of \(\text{arcsec}(x^2 + 1)\) results in \(\frac{2x}{(x^2 + 1)\sqrt{(x^2 + 1)^2 - 1}}\)

The formula might seem complex due to the absolute value and square root terms, but it's vital for capturing how changes in \(g(x)\) affect the function \(y\). It's especially useful when applying the product and chain rules in calculus.