Problem 23 - Brief Applied Calculus

Question

Find the derivative of each function. $$ g(x)=\sqrt{x}-\frac{1}{x} $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{x^2} \).

Step 1: Identify the Terms to Differentiate

First, notice that the function given is \( g(x) = \sqrt{x} - \frac{1}{x} \). This function is a difference of two terms: \( \sqrt{x} \) and \( \frac{1}{x} \). We need to find the derivative of each term separately.

Step 2: Rewrite the Function in Power Form

Before differentiating, it's easier if we rewrite \( \sqrt{x} \) and \( \frac{1}{x} \) in power form. We know that \( \sqrt{x} \) is the same as \( x^{1/2} \), and \( \frac{1}{x} \) is equivalent to \( x^{-1} \). So, the function can be rewritten as \( g(x) = x^{1/2} - x^{-1} \).

Step 3: Differentiate Each Term

Using the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \), we differentiate each term:- For \( x^{1/2} \), the derivative is \( \frac{1}{2}x^{-1/2} \).- For \( x^{-1} \), the derivative is \( -1x^{-2} \).

Step 4: Write the Derivative of the Function

Combine the derivatives of each term to find the derivative of the whole function. So, the derivative \( g'(x) \) is:\[ g'(x) = \frac{1}{2}x^{-1/2} - (-x^{-2}) \]Simplifying gives:\[ g'(x) = \frac{1}{2}x^{-1/2} + x^{-2} \]

Step 5: Rewrite in Radical Form (Optional)

If desired, rewrite the derivative from power form back to radical form:\[ g'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{x^2} \].

Key Concepts

Power Rule

The Power Rule is a fundamental principle in calculus for finding derivatives. It provides a straightforward method to differentiate functions of the form \( x^n \). The rule states that if you have a function \( f(x) = x^n \), then its derivative \( f'(x) \) is given by \( nx^{n-1} \). This means you multiply the power by the coefficient (which is 1 if not specified) and then reduce the exponent by one.

Let's apply this concept to each term of the function in the exercise. Consider \( g(x) = x^{1/2} - x^{-1} \):

• For the term \( x^{1/2} \), the power is \( 1/2 \). Applying the Power Rule: bring the power down as a multiplier \( \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} \).
• For the term \( x^{-1} \), the power is \( -1 \). Using the Power Rule: \( -1 \times x^{-1-1} = -x^{-2} \).

Thus, the Power Rule allows us to differentiate functions quickly and efficiently, turning exponents into coefficients and lowering the exponents.

Differentiation

Differentiation is a key concept in calculus that involves finding the derivative of a function. A derivative essentially measures how a function's output changes as its input changes. This is incredibly useful for understanding the rate of change or the slope of a function at any given point.

When differentiating the function given in the exercise, we dealt with separate terms to simplify the process. For a function like \( g(x) = \sqrt{x} - \frac{1}{x} \), it's easier to handle each part individually after rewriting it in power form. This separation is vital in calculus because it helps manage complex expressions.

After rewriting the function in power notation:

• We examined each term, utilizing the Power Rule to differentiate.
• Once each derivative was calculated individually as \( \frac{1}{2}x^{-1/2} \) and \( -x^{-2} \), we combined them to find the derivative of the entire function: \( g'(x) = \frac{1}{2}x^{-1/2} + x^{-2} \).

Differentiation is about breaking down functions into simpler parts to find their behavior and change dynamically through their derivatives.

Radical Functions

Radical functions involve roots, such as square roots or cube roots. These types of functions can often seem less straightforward to differentiate than polynomial functions because of their non-standard exponents. However, by transforming them into a power form, we can tackle differentiation easily using calculus rules.

In our original function \( g(x) = \sqrt{x} - \frac{1}{x} \), the \( \sqrt{x} \) term is a radical expression. To differentiate, we first expressed it in power form, which becomes \( x^{1/2} \). This conversion makes it manageable with the Power Rule.

• Radical functions like \( \sqrt{x} \) are just another form of exponential expressions \( x^{1/2} \).
• After transforming radical functions into a standard power form, they become suitable for standard differentiation techniques.

By converting radical terms to powers, differentiating such functions becomes straightforward and follows the same processes as typical polynomial differentiation. Thus, recalibrating radical functions into a power format is not only essential but also very handy for further calculus operations.