Problem 20 - Calculus Single Variable

Question

An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(\tan (x)\)

Step-by-Step Solution

Answer

First derivative: \( \sec^2(x) \); Second derivative: \( 2\sec^2(x)\tan(x) \); Third derivative: \( 4\sec^2(x)\tan^2(x) + 2\sec^4(x) \).

Step 1: Differentiate to find the First Derivative

To find the first derivative of the function, differentiate the expression \( f(x) = \tan(x) \). The derivative of \( \tan(x) \) with respect to \( x \) is \( \sec^2(x) \). Thus, \( f'(x) = \sec^2(x) \).

Step 2: Differentiate to find the Second Derivative

Now, differentiate \( f'(x) = \sec^2(x) \) to find the second derivative. Use the chain rule. The derivative of \( \sec^2(x) \) is \( 2\sec(x) \cdot \sec(x)\tan(x) = 2\sec^2(x)\tan(x) \). Thus, \( f''(x) = 2\sec^2(x)\tan(x) \).

Step 3: Differentiate to find the Third Derivative

Finally, differentiate \( f''(x) = 2\sec^2(x)\tan(x) \). Apply the product rule, which states that \((uv)' = u'v + uv'\). Here, \( u = 2\sec^2(x) \) and \( v = \tan(x) \). Then, \( u' = 4\sec^2(x)\tan(x) \) and \( v' = \sec^2(x) \). The third derivative is: \[ f'''(x) = (4\sec^2(x)\tan(x))\tan(x) + (2\sec^2(x))(\sec^2(x)) = 4\sec^2(x)\tan^2(x) + 2\sec^4(x). \]

Key Concepts

First Derivative

Calculus derivatives are an essential tool in understanding how functions change. The first derivative of a function provides us with the rate at which the function's value is changing at any given point. When we "take the derivative," we are essentially finding the slope of the tangent line to the curve of the function at that point.

Let's take a closer look using the example provided. Given the function \( f(x) = \tan(x) \), determining its first derivative means differentiating \( \tan(x) \) with respect to \( x \). Calculating this derivative involves using the differentiation rules known for trigonometric functions. Specifically, the derivative of \( \tan(x) \) is \( \sec^2(x) \). Hence, the first derivative, denoted as \( f'(x) \), is \( \sec^2(x) \).

This result tells us how steeply the function \( \tan(x) \) is increasing or decreasing at any point \( x \). The higher the value of \( \sec^2(x) \), the steeper the slope. When graphed, this illustrates the function's inclination at any particular point on its curve.

Second Derivative

The second derivative provides additional insight, indicating the "rate of change" of the first derivative itself. This is commonly referred to as the function's concavity. If the second derivative is positive, the function is concave up; if negative, concave down. The process to find it involves differentiating the first derivative again.

With \( f'(x) = \sec^2(x) \) from our earlier step, obtaining the second derivative involves differentiating \( \sec^2(x) \). By applying the chain rule for calculus—which allows us to differentiate composite functions—we find that the derivative of \( \sec^2(x) \) is \( 2\sec^2(x)\tan(x) \). Thus, the second derivative, \( f''(x) \), is equal to \( 2\sec^2(x)\tan(x) \).

This second derivative provides deeper insights into how \( \tan(x) \) behaves. By examining the value and sign of the second derivative, we can determine where the function changes its curvature, helping us to understand points of inflection and the overall shape of \( \tan(x) \) over its interval.

Third Derivative

Continuing with our derivative exploration, the third derivative offers even more depth. It examines the rate at which the second derivative is changing, shedding light on the function's behavior over time. In practical terms, while the first derivative gives the slope, and the second tells us about concavity, the third derivative highlights changes or trends in concavity.

For \( f''(x) = 2\sec^2(x)\tan(x) \), the third derivative requires applying the product rule. This differentiation rule helps with functions that are products of two functions—stated mathematically as \( (uv)' = u'v + uv' \). Applying this rule:

• Let \( u = 2\sec^2(x) \) and \( v = \tan(x) \).
• Then \( u' = 4\sec^2(x)\tan(x) \) and \( v' = \sec^2(x) \).

Plugging these into the product rule, the third derivative \( f'''(x) \) becomes:\[ 4\sec^2(x)\tan^2(x) + 2\sec^4(x) \]

This third derivative detects subtle shifts in how the function \( \tan(x) \) evolves. It’s particularly useful in advanced applications of calculus, such as in the analysis of dynamic systems where nuanced shifts in curvature could indicate significant future changes.