Problem 28 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\sqrt{x} \tan ^{3}(\sqrt{x}) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{3}{2} \cdot \frac{\tan^2(\sqrt{x})}{\sqrt{x}} \cdot \sec^2(\sqrt{x}) \)."

Step 1: Identify the Function Composition

The given function can be seen as a combination of two functions: an outer function and an inner function. Let's define these functions: - Outer function: \( u = an^3(v) \)- Inner function: \( v = \sqrt{x} \)Where \( y = u \), and we should find \( \frac{dy}{dx} \).

Step 2: Differentiate the Outer Function Using the Chain Rule

Using the chain rule, differentiate the outer function \( u = \tan^3(v) \) with respect to \( v \): \[ \frac{du}{dv} = 3\tan^2(v) \cdot \frac{d}{dv}(\tan(v)) = 3\tan^2(v) \cdot \sec^2(v) \]

Step 3: Differentiate the Inner Function

Differentiate the inner function \( v = \sqrt{x} \) with respect to \( x \):\[ \frac{dv}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \]

Step 4: Apply the Chain Rule to Find dy/dx

Now, using the chain rule, we combine the results from the previous steps to calculate \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = 3\tan^2(\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]Simplify this to obtain the final result.

Step 5: Simplify the Derivative Expression

The expression for \( \frac{dy}{dx} \) simplifies as follows:\[ \frac{dy}{dx} = \frac{3}{2} \cdot \frac{\tan^2(\sqrt{x})}{\sqrt{x}} \cdot \sec^2(\sqrt{x}) \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. If you have a function composed of two or more layers, the chain rule allows you to differentiate each layer separately and then combine the results.
For any functions \( u \) and \( v \), where \( u = f(v) \) and \( v = g(x) \), the chain rule formula is given by:

• \( \frac{du}{dx} = \frac{du}{dv} \times \frac{dv}{dx} \)

To apply the chain rule effectively:

• Identify the outer and inner functions first.
• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to \( x \).
• Multiply the derivatives together.

This rule is particularly useful when dealing with functions that have compositions involving powers, roots, trigonometric, exponential, or logarithmic functions.

Differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function is changing at any given point. This rate is mathematically represented by the derivative.
The process of differentiation can be applied to polynomial, exponential, logarithmic, and trigonometric functions to determine their derivatives.
Here are a few key points about differentiation:

• The derivative of a constant is zero.
• The derivative of \( x^n \) is \( nx^{n-1} \).
• Apply specific rules like the product, quotient, or chain rule when necessary.

Differentiation is essential in understanding how a function behaves, depicting its slope, and finding critical points like maxima, minima, and points of inflection. In our example, functions are nested and require careful application of rules to simplify derivatives accurately.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in various branches of mathematics and applied disciplines. These functions are periodic and relate angles in a right triangle to ratios of side lengths. In calculus, they have specific derivatives:

• \( \frac{d}{dx} \sin(x) = \cos(x) \)
• \( \frac{d}{dx} \cos(x) = -\sin(x) \)
• \( \frac{d}{dx} \tan(x) = \sec^2(x) \)

Understanding these derivatives is vital when differentiating more complex expressions involving trigonometric compositions. While applying the chain rule, we used the identity \( \frac{d}{dv} \tan(v) = \sec^2(v) \), significant for handling tasks involving trigonometric functions like the given exercise where we found the derivative of \( \tan^3(\sqrt{x}) \). This illustrates the interconnected use of the chain rule with differentiation principles to tackle trigonometric expressions efficiently.