Problem 4 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ h(t)=\frac{\sqrt{t-1}}{\sqrt{t+1}} $$

Step-by-Step Solution

Answer

\[ h'(t)= \frac{1}{2} \Big( \frac{1}{t+1} + \frac{1}{t-1} \Big) \]

Step 1: Simplify the function

Rewrite the given function in a more convenient form for differentiation. The function is To simplify, convert the square roots into exponent form: This can be written as: This gives: \[ h(t) = \frac{ (t-1)^{1/2} }{ (t+1)^{1/2} } = (t-1)^{1/2} (t+1)^{-1/2} \]

Step 2: Apply the Product Rule for differentiation

Recall the product rule for differentiation: For functions \( u(t) \) and \( v(t) \): The derivative is: \[ (uv)' = u'v + uv' \] In this case: Let \( u(t) = (t-1)^{1/2} \) and \( v(t) = (t+1)^{-1/2} \) The derivatives are: \( u'(t) \) and \( v'(t) \) We find each:\[\begin{align*} u'(t) & = \frac{d}{dt}(t-1)^{1/2} = \frac{1}{2}(t-1)^{-1/2} \ v'(t) & = \frac{d}{dt}(t+1)^{-1/2} = -\frac{1}{2}(t+1)^{-3/2} \end{align*}\] This leads to: \( h'(t)=(t-1)^{1/2} \cdot (-\frac{1}{2} (t+1)^{-3/2}) + (t+1)^{-1/2} \cdot \frac{1}{2} (t-1)^{-1/2} \)

Step 3: Simplify the result

Factor out common terms and simplify the derivative expression. t-h'(t) = (t-1)^{1/2} \cdot (-\frac{1}{2} (t+1)^{-3/2}) + (t+1)^{-1/2} \cdot \frac{1}{2} (t-1)^{-1/2} \( h'(t) \) simplifies to \[ (t-1)^{1/2} \cdot -\frac{1}{2} (t+1)^{-3/2} + (t+1)^{-1/2} \cdot \frac{1}{2} (t-1)^{-1/2} \]Combining like terms gives: \[-\frac{(t-1)^{1/2}}{2(t+1)^{3/2}} + \frac{(t+1)^{-1/2}}{2(t-1)^{1/2}} \]

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. It gives us crucial information about the behavior of the function, such as where it is increasing or decreasing and its concavity. To differentiate a function, we apply specific rules based on the function's form. For example, derivatives of polynomial functions are found using the power rule, which is straightforward but effective. Other rules, like the product rule and chain rule, help in dealing with more complex expressions.

Product Rule

The product rule is specifically useful when differentiating functions that are products of two or more functions. Suppose we have two differentiable functions, \(u(t)\) and \(v(t)\). The product rule states that the derivative of their product is: \( (uv)' = u'v + uv' \)To apply this correctly, we need to:

• Identify the two functions being multiplied together.
• Compute the derivative of each function separately.
• Use the product rule formula to find the derivative of the product.

In our exercise, we simplified \( h(t) \) to \( (t-1)^{1/2}(t+1)^{-1/2} \), so we used \( u(t) = (t-1)^{1/2} \) and \( v(t) = (t+1)^{-1/2} \). Calculating each derivative, we applied the product rule seamlessly to find the overall derivative.

Simplifying Expressions

After differentiating, expressions can often look complicated. Simplifying is about making these expressions as neat and easy to work with as possible. We use algebraic techniques like combining like terms, factoring out common factors, and simplifying fractions.In our exercise, after applying the product rule, our derivative looked complex. We had to factor out common terms and simplify:\( (t-1)^{1/2} \times \frac{-1}{2} (t+1)^{-3/2} + (t+1)^{-1/2} \times \frac{1}{2} (t-1)^{-1/2} \)Through combining like terms, we ended up with a simplified expression:\(-\frac{(t-1)^{1/2}}{2(t+1)^{3/2}} + \frac{(t+1)^{-1/2}}{2(t-1)^{1/2}} \)

Exponent Form

Understanding exponent form is key to simplifying and differentiating functions effectively. Exponent form expresses roots and powers in a way that makes differentiation rules easier to apply. For instance, we can express square roots as fractional exponents: \( \sqrt{x} = x^{1/2} \)This conversion allows us to use the power rule of differentiation seamlessly. In our exercise, rewriting \( \sqrt{t-1} \) and \( \sqrt{t+1} \) as \( (t-1)^{1/2} \) and \( (t+1)^{1/2} \) respectively enabled us to use the product rule more easily.Always convert square roots and other radicals to exponent form before proceeding with differentiation. It simplifies both the process and the final expressions considerably.