Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=e^{x} \arcsin x^{2} $$

Step-by-Step Solution

Answer

The derivative is \(2x \frac{e^x}{\sqrt{1-x^4}} + e^x \arcsin(x^2)\).

Step 1: Identify the Function and Strategy

The function given is a product of two functions: \(e^{x}\) and \(\arcsin(x^2)\). We'll use the product rule for differentiation, which states that \(D_x[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\), along with the chain rule.

Step 2: Differentiate First Function

Differentiate \(e^x\) with respect to \(x\). The derivative of \(e^x\) is \(e^x\).

Step 3: Differentiate Second Function Using Chain Rule

Differentiate \(\arcsin(x^2)\). Let \(u = x^2\). Then \(\frac{d}{dx} \arcsin(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\). So, \(\frac{du}{dx} = 2x\). Therefore, the derivative of \(\arcsin(x^2)\) is \(\frac{1}{\sqrt{1-x^4}} \cdot 2x\).

Step 4: Apply Product Rule

Now apply the product rule: \[ D_x [e^x \cdot \arcsin(x^2)] = e^x \cdot \frac{1}{\sqrt{1-x^4}} \cdot 2x + \arcsin(x^2) \cdot e^x. \]

Step 5: Simplify the Expression

Combine similar terms to simplify the expression: \[ D_x [e^x \cdot \arcsin(x^2)] = 2x \cdot \frac{e^x}{\sqrt{1-x^4}} + e^x \cdot \arcsin(x^2). \]

Key Concepts

Product Rule

When differentiating a product of two functions, we use the product rule. This is a fundamental technique in calculus that helps us find derivatives of expressions like products. The product rule states that if you have two functions multiplied together, say \( f(x) \) and \( g(x) \), then the derivative of their product is:

• \( D_x[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)

This rule allows us to take each part of the product one at a time. Differentiate the first function while leaving the second function unchanged. Then, do the reverse: leave the first function unchanged and differentiate the second function. For the problem at hand, \( y = e^x \cdot \arcsin(x^2) \), we apply the product rule by first taking the derivative of \( e^x \) and then \( \arcsin(x^2) \), combining these with the rule.

Chain Rule

The chain rule is another cornerstone of differentiation. It is used when differentiating compositions of functions, a common scenario in calculus.The chain rule states:

• If a function \( y = g(f(x)) \), then its derivative is \( g'(f(x)) \cdot f'(x) \)

This rule tells us how to handle nested functions by differentiating the "outer" function and multiplying by the derivative of the "inner" function.In the context of \( y = e^x \cdot \arcsin(x^2) \), the chain rule helps us differentiate \( \arcsin(x^2) \). By setting \( u = x^2 \), we first differentiate \( \arcsin u \), which is \( \frac{1}{\sqrt{1-u^2}} \), and then multiply by the derivative of \( u \), which is \( 2x \). Combining these gives us the derivative for the nested portion \( \arcsin(x^2) \).

Exponential Functions

Exponential functions are a unique category of functions characterized by a constant base raised to a variable exponent. The most common is the natural exponential function, \( e^x \), where "e" is a mathematical constant approximately equal to 2.718.One of the key properties of \( e^x \) is that its derivative is itself:

• \( \frac{d}{dx} e^x = e^x \)

This makes differentiating exponential functions straightforward and is particularly helpful when applying the product rule in problems involving \( e^x \).In our exercise, the function \( e^x \) plays a significant role by maintaining its form even after differentiation, simplifying calculations when combined with other functions, such as \( \arcsin(x^2) \).

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \arcsin(x) \), allow us to find angles based on trigonometric ratios. They are the inverse operations of regular trigonometric functions, useful in various mathematical applications.The differentiation of an inverse trigonometric function like \( \arcsin(x) \) involves:

• The derivative is \( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}} \)

In our context with \( \arcsin(x^2) \), we apply the chain rule to incorporate this definition into a more complex situation. The derivative becomes \( \frac{1}{\sqrt{1-x^4}} \cdot 2x \) after taking into account the inner function \( x^2 \).Understanding how inverse trigonometric derivatives work helps us tackle more intricate problems, combining rules and functions effectively.