Problem 68 - Calculus and its applications

Question

Differentiate each function. \(f(t)=\left(t^{5}+3\right) \cdot \frac{t^{3}-1}{t^{3}+1}\)

Step-by-Step Solution

Answer

Apply the product and quotient rules to find \( f'(t) = \frac{5t^{7} - 5t^{4}}{t^{3}+1} + \frac{6t^{7} + 18t^{2}}{(t^{3}+1)^{2}} \).

Step 1: Identify the Function Type

The function given is a product of two functions: \( u(t) = t^5 + 3 \) and \( v(t) = \frac{t^3 - 1}{t^3 + 1} \). We must use the product rule to differentiate this.

Step 2: Apply the Product Rule

The product rule states that if \( f(t) = u(t) \, v(t) \), then the derivative \( f'(t) = u'(t) \, v(t) + u(t) \, v'(t) \). We will differentiate \( u(t) \) and \( v(t) \) separately and then apply this formula.

Step 3: Differentiate \( u(t) = t^5 + 3 \)

Differentiate \( u(t) \) with respect to \( t \). Since \( u(t) = t^5 + 3 \), \( u'(t) = 5t^4 \), as the derivative of a constant is zero.

Step 4: Differentiate \( v(t) = \frac{t^3 - 1}{t^3 + 1} \) Using the Quotient Rule

The quotient rule states \( \left( \frac{p(t)}{q(t)} \right)' = \frac{p'(t) q(t) - p(t) q'(t)}{(q(t))^2} \) for \( p(t) = t^3 - 1 \) and \( q(t) = t^3 + 1 \). First, find \( p'(t) = 3t^2 \) and \( q'(t) = 3t^2 \). Then, apply the quotient rule:\[ v'(t) = \frac{(3t^2)(t^3 + 1) - (t^3 - 1)(3t^2)}{(t^3 + 1)^2} = \frac{3t^5 + 3t^2 - 3t^5 + 3t^2}{(t^3 + 1)^2} = \frac{6t^2}{(t^3 + 1)^2} \].

Step 5: Substitute Derivatives Back into the Product Rule

Now that we have \( u'(t) = 5t^4 \) and \( v'(t) = \frac{6t^2}{(t^3 + 1)^2} \), substitute back into the product rule:\[ f'(t) = (5t^4)\left(\frac{t^3 - 1}{t^3 + 1}\right) + (t^5 + 3)\left(\frac{6t^2}{(t^3 + 1)^2}\right) \].

Step 6: Simplify the Derivative Expression

Simplify each term in the derivative expression:1. \( (5t^4) \left(\frac{t^3 - 1}{t^3 + 1}\right) = \frac{5t^7 - 5t^4}{t^3 + 1} \).2. \( (t^5 + 3) \left(\frac{6t^2}{(t^3 + 1)^2}\right) = \frac{6t^7 + 18t^2}{(t^3 + 1)^2} \).Combine into a single fraction:\[ f'(t) = \frac{(5t^7 - 5t^4)(t^3 + 1) + (6t^7 + 18t^2)}{(t^3 + 1)^2} \].Carefully expand, combine like terms, and simplify further if possible.

Key Concepts

Product Rule

The product rule is a fundamental method used in calculus to find the derivative of a function that is the product of two functions. Here, we have two functions, \( u(t) = t^5 + 3 \) and \( v(t) = \frac{t^3 - 1}{t^3 + 1} \), multiplied together. To find the derivative of their product, we apply the product rule formula:- If \( f(t) = u(t) \, v(t) \), then the derivative \( f'(t) = u'(t) \, v(t) + u(t) \, v'(t) \).This rule dictates that we differentiate each function separately, multiply the derivative of the first by the second function, and add this to the product of the first function with the derivative of the second.

Quotient Rule

The quotient rule is crucial when dealing with derivatives of functions that are ratios, such as our function \( v(t) = \frac{t^3 - 1}{t^3 + 1} \).The quotient rule formula is:- \( \left( \frac{p(t)}{q(t)} \right)' = \frac{p'(t) \, q(t) - p(t) \, q'(t)}{(q(t))^2} \)In this case:- \( p(t) = t^3 - 1 \), which differentiates to \( p'(t) = 3t^2 \).- \( q(t) = t^3 + 1 \), which differentiates to \( q'(t) = 3t^2 \).Plug these into the quotient rule to find \( v'(t) \), simplifying the expression where possible.

Calculus Functions

Calculus functions can be intricate and fascinating because they often involve various differentiation rules, such as the product and quotient rules we've discussed. To differentiate complex expressions, it is essential to:

• Identify function types (product, quotient, etc.)
• Apply appropriate differentiation rules
• Simplify derivative expressions carefully

Understanding and practicing these steps allows for effectively tackling differentiated calculus functions, making this powerful tool accessible for solving real-world problems. Being systematic and careful ensures accuracy in your solutions.