Problem 60 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. $$ f(x)=\sqrt{x}\left(x^{4}-5 x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 5x^{3.5} - 6x^{1.5} \).

Step 1: Apply the Product Rule

To differentiate the function \( f(x) = \sqrt{x}(x^4 - 5x^2) \), recognize that it is a product of \( u = \sqrt{x} \) and \( v = x^4 - 5x^2 \). The product rule states that \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u = \sqrt{x} \)

To find \( u' \), use the power rule. \( u = x^{1/2} \), so \( u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

Step 3: Differentiate \( v = x^4 - 5x^2 \)

Apply the power rule to each term individually. The derivative of \( x^4 \) is \( 4x^3 \), and the derivative of \( -5x^2 \) is \( -10x \). Therefore, \( v' = 4x^3 - 10x \).

Step 4: Substitute into the Product Rule Formula

Substitute \( u' \), \( v \), and \( v' \) into the product rule formula: \[ f'(x) = \left(\frac{1}{2\sqrt{x}}\right)(x^4 - 5x^2) + (\sqrt{x})(4x^3 - 10x) \].

Step 5: Simplify the Expression

Simplify each term separately. The first term becomes: \[ \frac{x^4 - 5x^2}{2\sqrt{x}} = \frac{x^{7/2} - 5x^{3/2}}{2} \].The second term becomes: \[ 4x^{3.5} - 10x^{1.5} \].Combine to get: \[ f'(x) = \frac{x^{7/2} - 5x^{3/2}}{2} + 4x^{3.5} - 10x^{1.5} \].

Step 6: Finalize the Simplification

Combine the terms for the final expression. After combining like terms, the result is: \[ f'(x) = 5x^{3.5} - 6x^{1.5} \].

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are products of two or more functions. It offers a solution to the problem of finding the derivative of a product. If you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \cdot v(x) \). The derivative of this product is given as:\[(uv)' = u'v + uv'\]This formula is crucial as it breaks down the derivative into a manageable form by taking the derivative of one function while keeping the other unchanged and then doing the reverse.

• \( u' \) is the derivative of \( u \)
• \( v' \) is the derivative of \( v \)

In the exercise, the function is split into \( u = \sqrt{x} \) and \( v = x^4 - 5x^2 \). Using the product rule, we can differentiate the entire function efficiently. This technique is especially useful when applied to polynomials and rooted expressions in calculus.

Power Rule

The power rule is a simple, yet powerful tool in calculus for differentiating terms of the form \( x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) is:\[ \frac{d}{dx} x^n = nx^{n-1} \]This concept is straightforward; simply multiply the term by its power and reduce the power by one. The power rule is evident in the exercise when differentiating both parts of the product.For instance, \( u = x^{1/2} \) is differentiated to become \( u' = \frac{1}{2}x^{-1/2} \), applying the power rule for a root. Similarly, for \( v = x^4 - 5x^2 \), the derivatives are \( v' = 4x^3 - 10x \). It's worth noting how quickly the power rule allows differentiation, even when multiple terms are involved.

• Applicable to power functions
• Fast and efficient differentiation method

This rule is an essential part of a student's calculus toolkit, allowing rapid progress in problems involving exponents.

Calculus

Calculus, a branch of mathematics, plays a crucial role in solving complex problems involving change and motion. At its core, calculus focuses on two main concepts: differentiation and integration. Differentiation allows us to find how a function changes at any point, giving us the slope of the tangent line to the curve of a function.

• Understanding how variables change
• Applications in physics, engineering, and economics

In our given exercise, we've tackled a problem using differentiation, facilitated by the product rule and the power rule. These techniques exemplify how calculus can break down complex expressions into simpler components. This way, derivatives can be computed efficiently, revealing slopes and rates of change. Calculus is foundational to understanding many natural phenomena, providing insight into instantaneous rates of change. It empowers students to tackle real-life problems, making it an indispensable tool in various scientific disciplines.