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 \(D_x y = e^x \arcsin(x^2) + \frac{2xe^x}{\sqrt{1-x^4}}\).

Step 1: Identify the Differentiation Rule

The given function is of the form \(y = u imes v\) where \(u = e^x\) and \(v = \arcsin(x^2)\). We'll use the product rule for differentiation, which states: \((uv)' = u'v + uv'\).

Step 2: Differentiate \(u = e^x\)

Differentiate \(u = e^x\). Since the derivative of \(e^x\) with respect to \(x\) is \(e^x\), we have: \(u' = e^x\).

Step 3: Differentiate \(v = \arcsin(x^2)\)

For \(v = \arcsin(x^2)\), use the chain rule. The derivative of \(\arcsin(g(x))\) is \(\frac{1}{\sqrt{1-(g(x))^2}}\cdot g'(x)\). Here \(g(x) = x^2\), so \(g'(x) = 2x\). Thus, the derivative is \(v' = \frac{2x}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}}\).

Step 4: Apply the Product Rule

Using the product rule from Step 1, substitute \(u'\), \(v\), \(u\), and \(v'\) into \((uv)' = u'v + uv'\):\[ D_x y = e^x \cdot \arcsin(x^2) + e^x \cdot \frac{2x}{\sqrt{1-x^4}} \].

Step 5: Simplify if Necessary

The expression \(D_x y = e^x \cdot \arcsin(x^2) + \frac{2xe^x}{\sqrt{1-x^4}}\) is already simplified. There are no further simplifications possible given standard calculus operations.

Key Concepts

Product Rule

When faced with differentiating a product of two functions, the product rule is your go-to tool. Imagine you have two functions, say \(u(x)\) and \(v(x)\). The product rule states that the derivative of their product \(y = u \cdot v\) is:

• \((uv)' = u'v + uv'\)

This rule is essential when functions are multiplied, like in our case \(y = e^x \cdot \arcsin(x^2)\). Here, we identified \(u = e^x\) and \(v = \arcsin(x^2)\), making the differentiation process much smoother by breaking it into manageable parts.
It's important to understand each component: find \(u'\), then \(v'\), and finally apply the product rule by combining them back together. This method allows you to differentiate products of functions correctly and efficiently.
Using the product rule simplifies the differentiation process of complex expressions significantly.

Chain Rule

The chain rule is indispensable when differentiating compositions of functions. It's like peeling an onion—removing layer after layer. If you have a function \(f(g(x))\), the chain rule helps you find its derivative using:

• \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

In our example, we differentiate \(v = \arcsin(x^2)\). The outer function is the arcsine function \(\arcsin(g(x))\) and the inner function is \(g(x) = x^2\).
To apply the chain rule here:

• First, differentiate the outer function: the derivative of \(\arcsin(y)\) is \(\frac{1}{\sqrt{1-y^2}}\).
• Then, differentiate the inner function \(x^2\) to get \(2x\).
• Multiply these derivatives: \(\frac{2x}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}}\).

The chain rule thus enables differentiation of complex, nested functions, maintaining accuracy and efficiency.

Derivative of Exponential Functions

The exponential function \(e^x\) is one of the most interesting functions in calculus because its differential properties make it exceptionally straightforward. The key point to remember is:

• The derivative of \(e^x\) with respect to \(x\) is just \(e^x\) again.

This simplicity is due to the unique properties of the exponential function where the rate of change is proportional to the value of the function itself.
In our differentiation problem, with \(u = e^x\), applying the derivative directly gives \(u' = e^x\).
This enables us to handle exponential functions easily within compound functions like products and compositions, without added complexity. The exponential's self-replicating derivative makes it a crucial element in calculus, simplifying differential equations and models involving growth and decay.

Inverse Trigonometric Functions

Inverse trigonometric functions can be a bit tricky due to their domain restrictions and derivatives. In calculus, they often appear in differentiation problems, especially in compositions. For example, consider the inverse sine function \(\arcsin(x)\).
Its derivative is given by:

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

In our exercise, the function \(v = \arcsin(x^2)\) involves this derivative, but since \(x^2\) is inside arcsine, the chain rule needs to be applied as well. This leads to the overall expression:

• \(\frac{d}{dx}(\arcsin(x^2)) = \frac{2x}{\sqrt{1-x^4}}\)

Inverse trigonometric functions play a key role in calculus, especially in integrals and derivatives, helping describe angles and related measurements in various functions.