Problem 25 - Brief Applied Calculus

Question

Evaluate each expression. $$ \left.\frac{d^{2}}{d x^{2}} \sqrt{x^{3}}\right|_{x=1 / 16} $$

Step-by-Step Solution

Answer

The value of the second derivative at \( x = \frac{1}{16} \) is 3.

Step 1: Simplify the Expression under Differentiation

First, we simplify the expression inside the square root. The expression is \( \sqrt{x^3} \), which can be rewritten as \( x^{3/2} \). This puts it in a form that is easier to differentiate.

Step 2: Differentiate the Function First Time

To find the second derivative, we first need the first derivative of \( f(x) = x^{3/2} \). The first derivative is obtained using the power rule: \( f'(x) = \frac{3}{2}x^{1/2} \).

Step 3: Differentiate the Function Second Time

Now, we take the derivative of the first derivative \( f'(x) = \frac{3}{2}x^{1/2} \). Applying the power rule again, we get \( f''(x) = \frac{3}{4}x^{-1/2} \).

Step 4: Evaluate the Second Derivative at Given Point

Finally, we evaluate \( f''(x) \) at \( x = \frac{1}{16} \). Substitute \( x = \frac{1}{16} \) into \( f''(x) = \frac{3}{4}x^{-1/2} \):\[ f''\left(\frac{1}{16}\right) = \frac{3}{4} \left(\frac{1}{16}\right)^{-1/2}\]Simplify the term \( \left(\frac{1}{16}\right)^{-1/2} \) which is equal to 4, hence:\[ f''\left(\frac{1}{16}\right) = \frac{3}{4} \times 4 = 3\]

Step 5: Write the Final Answer

The value of the second derivative of \( \sqrt{x^3} \) at \( x = \frac{1}{16} \) is 3.

Key Concepts

Second Derivative

When dealing with calculus differentiation, a second derivative gives us even more detailed information about a function than the first derivative. It is the derivative of the derivative. Generally, the first derivative of a function helps us understand its rate of change or its slope. The second derivative, on the other hand, provides information about the curvature or concavity of the function.

In mathematical terms, if we have a function \( f(x) \), the first derivative is \( f'(x) \), and the second derivative is \( f''(x) \). Calculating this second derivative involves differentiating the first derivative. This process can give us insights into how the function behaves:

• If \( f''(x) > 0 \), the function is concave up, meaning it curves upwards.
• If \( f''(x) < 0 \), the function is concave down, which means it curves downwards.
• If \( f''(x) = 0 \), it might indicate a point of inflection, where the curvature changes direction.

Understanding these characteristics can be particularly useful in analyzing physical systems, optimizing problems, and predicting trends.

Power Rule

The power rule is a fundamental tool in calculus, especially useful when it comes to taking derivatives of polynomial expressions. It states that for any function \( f(x) = x^n \), the derivative \( f'(x) \) is found by multiplying by the exponent and then subtracting one from the exponent, represented as:

\( f'(x) = nx^{n-1} \).

For example, if \( n = 3 \), then the derivative of \( f(x) = x^3 \) becomes \( f'(x) = 3x^2 \). Similarly, for our given function \( f(x) = x^{3/2} \), the power rule allows us to differentiate quickly.

• First derivative: \( f'(x) = \frac{3}{2} x^{1/2} \).
• Second derivative: differentiate \( f'(x) \) again: \( f''(x) = \frac{3}{4}x^{-1/2} \).

By applying the power rule twice, we can smoothly transition from the original function to its second derivative, simplifying the process of finding how the curve changes.

Function Evaluation at a Point

After determining the second derivative, the exercise guides us to evaluate this derivative at a specific point. This part is known as "function evaluation at a point." It's particularly useful in gaining specific insights into a function's behavior at a particular value.

In our example, we find \( f''(x) = \frac{3}{4}x^{-1/2} \) and are instructed to evaluate it at \( x = \frac{1}{16} \). Evaluating involves substituting the specific \( x \) value into the second derivative:
\( f'' \left( \frac{1}{16} \right) = \frac{3}{4} \left( \frac{1}{16} \right)^{-1/2} \).

Simplifying this further, \( \left( \frac{1}{16} \right)^{-1/2} \) is equivalent to 4. So the computation becomes:
\( f'' \left( \frac{1}{16} \right) = \frac{3}{4} \times 4 = 3 \).

Through this process, we gain a concrete numerical value showing precisely how the function behaves at \( x = \frac{1}{16} \), marking how the curve bends at that point.