Problem 9 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ g(t)=\sqrt{2 t}+\sqrt{\frac{2}{t}} $$

Step-by-Step Solution

Answer

\( g'(t) = \frac{1}{\sqrt{2}} t^{-1/2} - \frac{1}{\sqrt{2}} t^{-3/2} \)

Step 1: Write the function in a simpler form

Rewrite the function in terms of exponents for easier differentiation: \[ g(t) = (2t)^{1/2} + \bigg(\frac{2}{t}\bigg)^{1/2} \]This can be further simplified as: \[ g(t) = 2^{1/2} t^{1/2} + 2^{1/2} t^{-1/2} \]

Step 2: Differentiate term by term

Use the power rule for differentiation, which states \( \frac{d}{dt}\big(t^n\big) = n t^{n-1} \), on each term separately: First term: \[ \frac{d}{dt}\big( 2^{1/2} t^{1/2} \big) = 2^{1/2} \cdot \frac{1}{2} t^{-1/2} \] Second term: \[ \frac{d}{dt}\big( 2^{1/2} t^{-1/2} \big) = 2^{1/2} \cdot ( -\frac{1}{2} ) t^{-3/2} \]

Step 3: Combine the results

Combine the derivatives of the individual terms: \[ g'(t) = 2^{1/2} \cdot \frac{1}{2} t^{-1/2} + 2^{1/2} \cdot ( -\frac{1}{2} ) t^{-3/2} \] Simplify the expression: \[ g'(t) = \frac{2^{1/2}}{2} t^{-1/2} - \frac{2^{1/2}}{2} t^{-3/2} \] \[ g'(t) = \frac{1}{\sqrt{2}} t^{-1/2} - \frac{1}{\sqrt{2}} t^{-3/2} \]

Key Concepts

Power Rule

The power rule is a basic rule in differentiation that helps us find the derivative of functions involving exponents. It's straightforward to use and powerful when dealing with polynomial expressions. The rule states that if you have a function of the form \(t^n\), then its derivative is given by:

\[ \frac{d}{dt}(t^n) = n t^{n-1} \]

This means we multiply by the current exponent and then reduce the exponent by one. For example, if \(t^4\) is the function, its derivative would be \(4t^3\). This process simplifies the differentiation of any polynomial function, making it manageable and systematic. In the given exercise, applying the power rule to each term separately simplifies finding the derivative.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative measures how a function changes as its input changes. In other words, it represents the rate of change or the slope of the curve at any given point.

To differentiate a function, you perform operations to calculate its derivative. For a sum of functions, like in our exercise with
\( g(t) = (2t)^{1/2} + \bigg(\frac{2}{t}\bigg)^{1/2} \),
we treat each part individually and then add the results. This principle of term-by-term differentiation keeps things simple and organized. By using rules like the power rule for each term and combining the results, you get the overall derivative.

It's vital to master differentiation rules as they form the foundation for more complex calculus topics.

Simplification of Exponents

Simplification of exponents makes complicated expressions easier to manage, especially when differentiating. To simplify, we convert roots and fractions into exponents. For instance,
\( \frac{2}{t} \) can be written as
\(2t^{-1}\), and \( \big(\frac{2}{t}\big)^{1/2} \) becomes \(2^{1/2} t^{-1/2}\). This makes applying the power rule straightforward.

Simplifying
\( g(t) = \big(2t\big)^{1/2} + \bigg(\frac{2}{t}\bigg)^{1/2} \) to
\( g(t) = 2^{1/2} t^{1/2} + 2^{1/2} t^{-1/2} \),
shows clear terms to differentiate.
After differentiation, recombining into a simplified form, like
\( g'(t) = \frac{1}{\text{2}} \frac{t^{-1/2}}{\text{}} - \frac{1}{2} \big( t^{-3/2} \big) \),
leaves us with a simple and understandable result.

Mastering this allows easy manipulation of terms and streamlining the differentiation process, resulting in cleaner and more interpretable results.