Question

Compute the derivative of the given function. $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}$$

Short answer

The derivative of the function is \( f'(x) = \frac{1}{2}x^{-3/2}(x - 1) \).

Step-by-step solution

Rewrite the Function

First, we can rewrite the function in a more differentiable form. Instead of using a square root, express square roots as fractional exponents. The given function is \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} \). This can be rewritten as \( f(x) = x^{1/2} + x^{-1/2} \).

Differentiate the Function Using Power Rule

Now, apply the power rule for differentiation which states that \( \frac{d}{dx}(x^n) = nx^{n-1} \). Differentiate \( f(x) = x^{1/2} + x^{-1/2} \): - For \( x^{1/2} \), the derivative is \( \frac{1}{2}x^{-1/2} \).- For \( x^{-1/2} \), the derivative is \( -\frac{1}{2}x^{-3/2} \).

Combine Derivatives

Combine the derivatives obtained from both terms. Therefore, the derivative \( f'(x) \) is:\[ f'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \].

Simplify Expression

The expression can be simplified if needed by factoring out common terms. Here, both terms share a common factor of \( \frac{1}{2}x^{-3/2} \). Factoring this out gives:\[ f'(x) = \frac{1}{2}x^{-3/2}(x - 1) \].

Key concepts

Power Rule

The power rule is a cornerstone technique in calculus used for finding the derivative of polynomial functions. It states that if you have a term \( x^n \), its derivative is computed as \( nx^{n-1} \). This rule simplifies the process of differentiation and allows you to handle a wide range of expressions quickly.
To apply the power rule:

• Identify the power of the variable, which is the exponent \( n \) in the term \( x^n \).
• Multiply the entire term by this power \( n \), thus transforming it into \( nx^n \).
• Decrease the power by one, resulting in \( nx^{n-1} \) as the final derivative.

In this exercise, we start by rewriting \( \sqrt{x} \) as \( x^{1/2} \) and \( \frac{1}{\sqrt{x}} \) as \( x^{-1/2} \), then apply the power rule to find the derivatives of each part, which results in \( \frac{1}{2}x^{-1/2} \) and \(-\frac{1}{2}x^{-3/2} \), respectively.
This approach is simple and effective, especially for terms that might initially appear intimidating in a different form. Mastering this rule significantly enhances one's ability to differentiate complex functions in calculus.

Fractional Exponents

Fractional exponents are a way of expressing roots, such as square or cube roots,in a form that makes algebraic manipulation easier. Instead of writing \( \sqrt{x} \), you can express it as \( x^{1/2} \). This is particularly useful in differentiation and integration, as the rules for exponents apply consistently even when they are fractions.
When handling fractional exponents:

• Remember that the numerator represents the power, and the denominator indicates the root.
• For example, \( x^{3/2} \) means you take the square root of \( x \) and then raise it to the power of 3.
• These expressions allow us to apply calculus rules like the power rule easily, making the process less cumbersome.

In our function \( f(x) = x^{1/2} + x^{-1/2} \), translating to fractional exponents makes applying the power rule straightforward.
By doing this transformation, we can smoothly differentiate the function without the need to manipulate cumbersome root expressions repeatedly.
This steams up the process and simplifies it beautifully.

Differentiation

Differentiation is the process through which we calculate the rate at which a function changes at any given point. It's the primary operation in differential calculus. Differentiation can help determine slopes of curves and optimize various functions in mathematical modeling.
To differentiate successfully:

• First, express the function in a suitable form if it isn't already (using tools like fractional exponents).
• Apply the relevant differentiation rules like the power rule for each term in the function.
• Combine the resulting terms to get the full derivative of the function.

Differentiation is central to solving problems involving motion, growth, and change -in essence, it models real-world scenarios very effectively.
In this problem, differentiation allows us to move from \( f(x) = x^{1/2} + x^{-1/2} \) to a derivative \( f'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \), which constitutes the instantaneous rate of change for every point in the original function.
This derivative serves as a powerful tool in predicting how the function behaves overall.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them easier to work with. This skill becomes essential when preparing functions for differentiation or integration. It often involves processes like rewriting terms, factoring, expanding, and combining like terms.
In this problem, algebraic manipulation allowed us to refine the derivative from \( f'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \) into a simplified form by factoring out common terms.

• Recognize common factors in the terms, in this case, \( \frac{1}{2}x^{-3/2} \).
• Factor out this common component to simplify the expression, resulting in \( \frac{1}{2}x^{-3/2}(x - 1) \).
• Such simplifications make further mathematical operations tidier and often more intuitive.

Through smart algebraic manipulation, we can transform complex derivatives into straightforward expressions that are much easier to evaluate or utilize in further calculations. It's a critical step in the calculus toolkit, enhancing both understanding and computational efficiency.