Problem 50 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ g(t)=\log _{2}\left(t^{2}+7\right)^{3} $$

Step-by-Step Solution

Answer

The derivative of the function \(g(t)=\log _{2}\left(t^{2}+7\right)^{3}\) is \(g'(t) = 6t/((t^2 + 7) \ln 2)\).

Step 1: Simplification

Utilize the rules of logarithm to simplify the function. The rule \(\log_b(x^y) = y\log_b(x)\) will be especially useful. The function \( g(t) = \log_2\left(t^2 + 7\right)^3 \) will become \( g(t) = 3\log_2\left(t^2 + 7\right) \).

Step 2: Differentiating with respect to t

Next step consists of taking derivative of the simplified function. The derivative of a log function based on the power rule \( \log_a(u)' = 1/(u \ln a) \cdot u' \) can be applied. For \(u = t^2 + 7\), the derivative \(u' = 2t\). Substituting these into the rule will give \(g'(t) = 3 [1/((t^2 + 7) \ln 2)] \cdot 2t \).

Step 3: Simplifying the Derivative

Now, simplify this answer to get the final derivative of the function. The derivative simplifies to \(g'(t) = 6t/((t^2 + 7) \ln 2)\).

Key Concepts

Logarithmic Properties

Logarithmic properties are incredibly useful when dealing with functions that include logs raised to powers, as seen in our function, \( g(t) = \log_2\left(t^2+7\right)^3 \). Understanding and applying these properties allows us to simplify complex expressions, making the differentiation process much easier.

One key property is the power rule for logarithms, which states:
- \( \log_b(x^y) = y\log_b(x) \).
This property helps us turn the exponent into a coefficient, simplifying the differentiation step.

In this exercise, the expression \( \log_2\left(t^2+7\right)^3 \) can be rewritten as \( 3\log_2(t^2+7) \), thanks to the logarithmic power rule. This simplification makes the next steps of finding the derivative more straightforward.

Differentiation

Once the function is simplified using logarithmic properties, differentiation is the next step. Differentiation is the process of finding the derivative of a function, which represents the rate at which the function changes with respect to its variable.

For the function \( g(t) = 3\log_2(t^2+7) \), we can apply the derivative rule for logarithmic functions. The general formula is:

• \( \frac{d}{dt} \log_a(u) = \frac{1}{u \ln a} \cdot \frac{du}{dt} \)

Here, \( u = t^2 + 7 \), so \( \frac{du}{dt} = 2t \). By substituting these into the derivative formula, we find:

• \( g'(t) = 3 \cdot \left(\frac{1}{(t^2+7) \ln 2}\right) \cdot 2t \)

This computation is essential for understanding how quickly \( g(t) \) changes at any given point \( t \).

Simplifying Derivatives

After applying differentiation, we often have an expression that can be further simplified. Simplifying derivatives not only makes them easier to interpret but also prepares them for practical use in applications, like solving real-world problems involving rates of change.

In the current example, we've reached the expression \( g'(t) = 3 \cdot \frac{2t}{(t^2+7) \ln 2} \). Simplifying this involves combining constants and rearranging terms:

• Combine 3 and 2t to get \( 6t \).
• Place \( 6t \) over the denominator \((t^2+7) \ln 2\) directly.

This gives us the final simplified derivative: \( g'(t) = \frac{6t}{(t^2+7) \ln 2} \). With this process, the function derivative is now easy to use for calculating changes in \( g(t) \) relative to changes in \( t \).