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Natural logarithms are the logarithms to the base of the number e, where e is approximately 2.71828. The natural logarithm is often denoted as \( \ln \). It's used to transform multiplicative relationships into additive ones. This property is advantageous when dealing with more complex functions like expressions under a square root or exponents.
In logarithmic differentiation, taking the natural logarithm simplifies the expression by turning products into sums, and powers into multipliers. For example, for a function like \( y = \left(\frac{1}{t(t+1)}\right)^{1/2} \), applying a natural logarithm converts it into \( \ln(y) = \frac{1}{2} \ln\left(\frac{1}{t(t+1)}\right) \), allowing for easier differentiation.
By dealing with exponents and products separately after applying the natural logarithm, you can make the differentiation process much smoother.

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When a function is composed of two or more functions, the chain rule helps us find its derivative. For example, if you have a function \( u(t) \) inside another function, \( f(u) \), the chain rule states that the derivative \( \frac{d}{dt}[f(u(t))] = f'(u) \cdot u'(t) \).
In our exercise, consider the expression \( \ln(y) = -\frac{1}{2} ( \ln(t) + \ln(t+1) ) \). By the chain rule, the derivative \( \frac{d}{dt}(\ln(y)) \) involves not just computing the derivative of the natural log but also the derivative of the inside functions \( \ln(t) \) and \( \ln(t+1) \):

• \( \frac{d}{dt}(\ln(t)) = \frac{1}{t} \)
• \( \frac{d}{dt}(\ln(t+1)) = \frac{1}{t+1} \)

This technique breaks down complex derivative calculations into simpler parts by multiplying the derivatives of the outer and inner functions.

Exponents are used to express repeated multiplication of a number by itself. The properties of exponents make them very powerful tools in mathematics. Here are a few key properties:

• The exponent \((a^b)^c = a^{b \cdot c}\)
• Reciprocal: \(\frac{1}{a^b} = a^{-b}\)
• Product property: \(a^b \cdot a^c = a^{b+c}\)

These properties help in rewriting complex expressions into manageable forms.
In the exercise, the function \( \sqrt{\frac{1}{t(t+1)}} \) is rewritten using exponents as \( \left(\frac{1}{t(t+1)}\right)^{1/2} \). Using exponent properties, this can be further broken down as \( t^{-1/2}\cdot(t+1)^{-1/2} \).
This transformation is crucial for leveraging other calculus techniques like logarithmic differentiation by simplifying expressions for easier manipulation.

The product rule is a calculus technique used to find the derivative of a product of two functions. If you have two functions \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is \( (uv)' = u'v + uv' \).
While the product rule is not directly applied in this particular exercise because we use logarithmic differentiation, understanding it is essential in calculus. When dealing with products inside functions like \( t(t+1) \), simplifying them using logarithms first avoids the need to apply the product rule.
The derivative, in this case, is found without using the product rule, because the natural logarithm and properties of exponents effectively turn a product into a sum of simpler functions. This is why the combination of these techniques is so powerful in breaking down the function into easily differentiated components.