Problem 37 - Brief Applied Calculus

Question

31-38. Find the indicated derivatives. If \(f(x)=\frac{16}{\sqrt{x}}+8 \sqrt{x}\), find \(\left.\frac{d f}{d x}\right|_{x=4}\)

Step-by-Step Solution

Answer

The derivative evaluated at \(x = 4\) is 1.

Step 1: Express the Function in Power Form

The given function is \(f(x) = \frac{16}{\sqrt{x}} + 8\sqrt{x}\). Express each term with exponents: \(f(x) = 16x^{-1/2} + 8x^{1/2}\).

Step 2: Differentiate the Function

Differentiate each term separately. For \(16x^{-1/2}\), use the power rule: the derivative is \(16 \cdot (-1/2)x^{-1/2 - 1} = -8x^{-3/2}\). For \(8x^{1/2}\), the derivative is \(8 \cdot (1/2)x^{1/2 - 1} = 4x^{-1/2}\). The derivative of \(f(x)\) is \(f'(x) = -8x^{-3/2} + 4x^{-1/2}\).

Step 3: Evaluate the Derivative at x=4

Substitute \(x = 4\) into the derivative \(f'(x) = -8x^{-3/2} + 4x^{-1/2}\). Calculate each term: \(-8 \cdot 4^{-3/2} + 4 \cdot 4^{-1/2} = -8 \cdot \frac{1}{8} + 4 \cdot \frac{1}{2}\). This simplifies to \(-1 + 2 = 1\).

Step 4: Finalize the Result

Having evaluated the derivative at \(x = 4\), the result is \(f'(4) = 1\).

Key Concepts

Power Rule

The power rule is a crucial concept in calculus for finding derivatives. It's a simple yet powerful shortcut for differentiating functions of the form \(x^n\), where \(n\) is any real number. The power rule states that if \(f(x) = x^n\), then the derivative \(f'(x)\) is \(nx^{n-1}\). This means you take the exponent \(n\), multiply it by the coefficient in front of \(x\), and then decrease the exponent by 1.
In the original exercise, we transform the given function \(f(x) = \frac{16}{\sqrt{x}} + 8\sqrt{x}\) into a format where we can apply the power rule easily. By expressing the function in power form, \(x\) raised to a negative or fractional power, we have \(f(x) = 16x^{-1/2} + 8x^{1/2}\). This makes it straightforward to apply the power rule separately to each term:

• For \(16x^{-1/2}\), the derivative becomes \(-8x^{-3/2}\). Here, \(-1/2\) is multiplied by 16, and the exponent \(-1/2\) is decreased by 1.
• For \(8x^{1/2}\), the derivative becomes \(4x^{-1/2}\). Here, \(1/2\) is multiplied by 8, and the exponent \(1/2\) is decreased by 1.

The power rule simplifies what could otherwise be a complex task into a straightforward calculation.

Differentiation

Differentiation is the core process in calculus used to compute the derivative of a function. Derivatives tell us how functions change, and they have numerous applications from calculating rates of change to finding slopes of curves.
In the problem we are solving, differentiation is used on the function \(f(x) = 16x^{-1/2} + 8x^{1/2}\). The derivative, denoted as \( f'(x) \), gives us a new function that describes the rate of change of \(f(x)\) at any point \(x\). Differentiation requires applying rules like the power rule (which we've already detailed) to each term in the function individually. This results in the complete derivative:

• Applying the derivative rules, \(f'(x) = -8x^{-3/2} + 4x^{-1/2}\).

The process of differentiation can initially seem complex, but once you practice applying these rules, the process becomes intuitive. The ability to transform expressions and apply rules such as the power rule helps greatly in understanding and performing differentiation.

Evaluating Derivatives at a Point

Evaluating a derivative at a specific point involves substituting a given \(x\) value into the derived function \(f'(x)\). It's a simple yet important process that provides the rate of change of the function at that particular point.
For the problem at hand, once we've found the derivative, \(f'(x) = -8x^{-3/2} + 4x^{-1/2}\), we evaluate it at \(x = 4\). This involves:

• Substituting \(x = 4\) into each term of the derivative.
• Calculating with simplification, the expression \(-8 \cdot 4^{-3/2} + 4 \cdot 4^{-1/2}\).
• This gives \(-1 + 2 = 1\).

Evaluating the derivative tells us that at \(x = 4\), the rate of change of \(f(x)\) is 1. This insight can be exceptionally valuable, for instance, in physics for determining velocity at a particular instant or in economics for calculating marginal cost.