Problem 31 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ h(x)=x \tan (2 \sqrt{x})+7 $$

Step-by-Step Solution

Answer

The derivative of the function is \( h'(x) = \sqrt{x} \sec^2(2\sqrt{x}) + \tan(2\sqrt{x}) \).

Step 1: Differentiate the outer function

The given function is \( h(x) = x \tan(2\sqrt{x}) + 7 \). To find the derivative, start with differentiating the outer function with respect to \( x \). Use the product rule, which states that the derivative of a product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \).Let \( u(x) = x \) and \( v(x) = \tan(2\sqrt{x}) \), so \( u'(x) = 1 \) and \( v(x) = \tan(2\sqrt{x}) \).Using the product rule:\[ \frac{d}{dx}[x \tan(2 \sqrt{x})] = x \frac{d}{dx}[\tan(2\sqrt{x})] + 1 \cdot \tan(2\sqrt{x}). \]

Step 2: Differentiate the inner function

Now, determine \( \frac{d}{dx}[\tan(2\sqrt{x})] \). This requires using the chain rule because \( \tan(2\sqrt{x}) \) is a composition of functions.First,\( \frac{d}{du} \tan u = \sec^2(u) \). Here, let \( u = 2\sqrt{x} \).Next, we need the derivative of \( 2\sqrt{x} \) with respect to \( x \):\[ \frac{d}{dx}(2\sqrt{x}) = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = x^{-1/2} = \frac{1}{\sqrt{x}}. \]

Step 3: Apply the chain rule

Now apply the chain rule for differentiation:\[ \frac{d}{dx}[\tan(2\sqrt{x})] = \sec^2(2\sqrt{x}) \cdot \frac{1}{\sqrt{x}}. \]So, substituting back into the product rule:\[ \frac{d}{dx}[x \tan(2\sqrt{x})] = x \cdot \sec^2(2\sqrt{x}) \cdot \frac{1}{\sqrt{x}} + \tan(2\sqrt{x}). \]Simplify this expression:\[ \frac{d}{dx}[x \tan(2\sqrt{x})] = \sqrt{x} \cdot \sec^2(2\sqrt{x}) + \tan(2\sqrt{x}). \]

Step 4: Differentiate the constant

The derivative of the constant 7 is zero:\[ \frac{d}{dx}(7) = 0. \]

Step 5: Combine the results

Now, combine everything to get the derivative of \( h(x) \):\[ h'(x) = \sqrt{x} \sec^2(2\sqrt{x}) + \tan(2\sqrt{x}). \]

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. It simplifies the process by breaking down a challenging task into smaller, manageable parts. If you have two functions, say \( u(x) \) and \( v(x) \), the product rule tells us how to differentiate their product. The formula is:

• \( (uv)' = u'v + uv' \).

This means you take the derivative of the first function and multiply it by the second function as is, then add the undifferentiated first function multiplied by the derivative of the second function.
In the given exercise, you applied the product rule to \( h(x) = x \tan(2\sqrt{x}) + 7 \), treating \( x \) as \( u(x) \) and \( \tan(2\sqrt{x}) \) as \( v(x) \). You differentiated \( x \) to get \( 1 \) and then focused on \( \tan(2\sqrt{x}) \), which required further differentiation using the chain rule.

Chain Rule

The chain rule is an essential differentiation technique used when dealing with composite functions, or functions within functions. If you have a function \( y = f(g(x)) \), the chain rule helps you find the derivative by multiplying the derivative of \( f \) with respect to \( g \) by the derivative of \( g \) with respect to \( x \):

• \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \).

During the exercise, you found the derivative of \( \tan(2\sqrt{x}) \). You considered \( \tan(u) \) with \( u = 2\sqrt{x} \). First, you derived \( \tan(u) \) as \( \sec^2(u) \), then multiplied by the derivative of \( u = 2\sqrt{x} \), which is \( \frac{1}{\sqrt{x}} \). The chain rule's application allowed you to accurately differentiate this complex inner function.
This process is crucial when dealing with layered functions like \( \tan(2\sqrt{x}) \), ensuring precision in differentiation.

Differentiation

Differentiation is a process in calculus that allows us to determine the rate at which a function is changing. It is the fundamental concept behind deriving functions and relies on various rules, such as the product and chain rules, for efficiency.

• The notation \( \frac{d}{dx} \) represents the derivative of a function.
• A derivative provides the slope of the function at any given point, reflecting how fast or slow the changes occur.

In the exercise, differentiation was used to find \( h'(x) \) for the given function \( h(x) = x \tan(2\sqrt{x}) + 7 \). You differentiated each component, including the product \( x\tan(2\sqrt{x}) \) and the constant \( 7 \), whose derivative is zero. Each of these steps reflected various differentiation techniques, ensuring an accurate measure of the function's behavior over its domain.
By applying differentiation rules methodically, you maintain the structural integrity of the function while analyzing its precise changes.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a pivotal role in calculus, especially when it comes to differentiation. These functions describe relationships in triangles and oscillatory behavior across various domains, such as engineering and physics.

• In differentiation, \( \tan(x) \) becomes \( \sec^2(x) \).
• Trigonometric derivatives have their unique forms, crucial for accurately handling changes within trigonometric expressions.

For the given function \( h(x) = x \tan(2\sqrt{x}) + 7 \), \( \tan(2\sqrt{x}) \) required differentiation as part of the product and chain rule applications. Recognizing the derivative of \( \tan \) is \( \sec^2 \) ensures correctness in handling such expressions.
Understanding how each trigonometric function transforms through differentiation simplifies tackling these problems efficiently. Trigonometric differentiation requires familiarity with these periodic functions' unique characteristics and behaviors.