Problem 28 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

Use the product rule and simplify to obtain \(\frac{dy}{dx} = \frac{1}{2} x^{-1/2} \tan^3(\sqrt{x}) + \frac{3}{2} \tan^2(\sqrt{x}) \sec^2(\sqrt{x})\).

Step 1: Recognize the Functions Involved

The given function is a product of two functions: \(u(x) = \sqrt{x}\) and \(v(x) = \tan^3(\sqrt{x})\). We will need to differentiate this using the product rule.

Step 2: Apply the Product Rule

The product rule states that the derivative of a product \(u(x)\cdot v(x)\) is given by \(u'(x)v(x) + u(x)v'(x)\). We need to calculate \(u'(x)\) and \(v'(x)\) separately.

Step 3: Differentiate \(u(x) = \sqrt{x}\)

The derivative of \(u(x) = x^{1/2}\) is \(u'(x) = \frac{1}{2}x^{-1/2}\).

Step 4: Differentiate \(v(x) = \tan^3(\sqrt{x})\)

First, notice that \(v(x)\) can be written as \((\tan(\sqrt{x}))^3\). Using the chain rule, the derivative \(v'(x)\) is calculated by first deriving the outer function and then multiplying by the derivative of the inner functions: \[ v'(x) = 3(\tan(\sqrt{x}))^2 \cdot \sec^2(\sqrt{x}) \cdot (\frac{1}{2}x^{-1/2}) \]

Step 5: Combine Derivatives Using Product Rule

Substitute back into the product rule formula: \[ \frac{dy}{dx} = \left(\frac{1}{2}x^{-1/2}\right)\tan^3(\sqrt{x}) + \sqrt{x} \cdot 3(\tan(\sqrt{x}))^2 \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2}x^{-1/2} \]

Step 6: Simplify the Expression

Further simplification leads to: \[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2}\tan^3(\sqrt{x}) + \frac{3}{2}x^{-1/2} \cdot \tan^2(\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \sqrt{x} \] Which simplifies to:\[ \frac{dy}{dx} = \frac{1}{2} x^{-1/2} \tan^3(\sqrt{x}) + \frac{3}{2} \tan^2(\sqrt{x}) \sec^2(\sqrt{x}) \]

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus used to differentiate composite functions. A composite function is when one function is nested inside another. For example, if you have \( y = f(g(x)) \), the chain rule helps in finding the derivative of this composite function. The rule states that the derivative of \( y \) with respect to \( x \) is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). In simpler terms, you take the derivative of the outer function, keep the inside the same, and multiply it with the derivative of the inside. This means:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

• Use the chain rule when you encounter functions like \( \tan^3(\sqrt{x}) \), where \( \tan \) is applied to another function \( \sqrt{x} \).
• Remember to differentiate the outside function first (\( \tan^3 \)), then work your way inwards.
• This approach helps in tackling complex functions that seem intimidating at first glance.

Trigonometric Functions

Trigonometric functions such as \( \sin \), \( \cos \), and \( \tan \) are foundational in understanding periodic phenomena. In calculus, these functions have unique derivatives that are pivotal when solving problems.For the tangent function, \( \tan(x) \), the derivative is:\[ \frac{d}{dx} [\tan(x)] = \sec^2(x) \]When you have a function like \( \tan^3(\sqrt{x}) \), you:

• Differentiate the power part first (using exponent rules), which introduces a multiplier (here it is 3).
• Apply the chain rule to account for the \( \sqrt{x} \) inside the tangent.

Understanding these steps is crucial when dealing with derivatives that involve trigonometric expressions, as they often appear in oscillatory or wave-like motions in real-world contexts.

Derivatives

Derivatives measure how a function changes as its input changes. They provide valuable information about the slope of a function at any given point. In our exercise, you dealt with the function:\[ y = \sqrt{x} \tan^3(\sqrt{x}) \]To find its derivative \( \frac{dy}{dx} \), we utilized both the product rule and the chain rule. Here's a breakdown:

• The product rule allows you to differentiate two multiplied functions, requiring the derivative of each function individually before combining them.
• In this particular problem, first differentiate \( \sqrt{x} \), producing \( \frac{1}{2} x^{-1/2} \).
• Then, use the chain rule for the trigonometric part, following the nested differentiation process described earlier.

In essence, derivatives are a foundational tool in calculus for analyzing changes and forming the basis of more advanced topics like integrals and differential equations. Understanding the process behind finding derivatives encourages deeper mathematical insight and problem-solving strategies.