Problem 36 - Calculus and its applications

Question

Differentiate each function. \(G(t)=\left(3 t^{5}-t^{2}\right)\left(t-\frac{5}{t}\right)\)

Step-by-Step Solution

Answer

The derivative of the function is \( G'(t) = 18t^5 - 60t^3 - 3t^2 + 5 \).

Step 1: Identify the Product Rule

To differentiate the given function, observe that it is a product of two functions: \[ G(t) = u(t) imes v(t) \]where \( u(t) = 3t^5 - t^2 \) and \( v(t) = t - \frac{5}{t} \).The product rule for differentiation is given by \[ \frac{d}{dt}[u(t) \cdot v(t)] = u'(t)v(t) + u(t)v'(t) \].

Step 2: Differentiate u(t)

Differentiate \( u(t) = 3t^5 - t^2 \):The derivative of \( 3t^5 \) is \( 15t^4 \), and the derivative of \( -t^2 \) is \( -2t \).Thus, \( u'(t) = 15t^4 - 2t \).

Step 3: Differentiate v(t)

Differentiate \( v(t) = t - \frac{5}{t} \):The derivative of \( t \) is \( 1 \), and the derivative of \( -\frac{5}{t} \) is\( 5t^{-2} \) or equivalently \( \frac{5}{t^2} \).Thus, \( v'(t) = 1 + \frac{5}{t^2} \).

Step 4: Apply the Product Rule

Substitute \( u'(t) \), \( v(t) \), \( u(t) \), and \( v'(t) \) into the product rule:\[ G'(t) = (15t^4 - 2t)(t - \frac{5}{t}) + (3t^5 - t^2)(1 + \frac{5}{t^2}) \].

Step 5: Simplify Each Term

Simplify the expressions from the product rule application:For \( (15t^4 - 2t)(t - \frac{5}{t}) \), distribute each term:- \( 15t^4 \cdot t = 15t^5 \)- \( 15t^4 \cdot -\frac{5}{t} = -75t^3 \)- \( -2t \cdot t = -2t^2 \)- \( -2t \cdot -\frac{5}{t} = 10 \)For \( (3t^5 - t^2)(1 + \frac{5}{t^2}) \):- \( 3t^5 \cdot 1 = 3t^5 \)- \( 3t^5 \cdot \frac{5}{t^2} = 15t^3 \)- \( -t^2 \cdot 1 = -t^2 \)- \( -t^2 \cdot \frac{5}{t^2} = -5 \).

Step 6: Combine and Simplify

Combine like terms from both simplified expressions:Group the results:- \( 15t^5 + 3t^5 = 18t^5 \)- \( -75t^3 + 15t^3 = -60t^3 \)- \( -2t^2 - t^2 = -3t^2 \)- \( 10 - 5 = 5 \)Thus, the derivative is:\[ G'(t) = 18t^5 - 60t^3 - 3t^2 + 5 \].

Key Concepts

Understanding the Product Rule

The product rule is a fundamental tool in calculus differentiation used when differentiating expressions that involve the multiplication of two functions.
When you have a function expressed as the product of two separate functions, like in our problem where the function is expressed as \( G(t) = u(t) \cdot v(t) \), the product rule is indispensable.
The rule states that if you need to differentiate a product of two functions, you must take the derivative of each function individually and then apply the formula:

• First, differentiate the first function \( u(t) \)
• Multiply it by the second function \( v(t) \) as it is
• Then, add the product of the first function \( u(t) \) and the derivative of the second function \( v(t) \)

The product rule can be neatly expressed in the formula: \( \frac{d}{dt}[u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t) \). This rule lets us break down complex problems into simpler parts, each of which can be differentiated using basic derivative rules.

Mastering Function Derivatives

Function derivatives represent the rate of change of a function with respect to a variable. In simpler terms, it measures how a function changes as its input changes.
When working with derivatives, it's important to understand the fundamental rules of differentiation, which allow us to find these derivatives efficiently. Here are a couple of basic rules used in our exercise:

• Power Rule: If you have a function \( f(t) = at^n \), its derivative \( f'(t) \) is \( a \cdot nt^{n-1} \).
• Constant Rule: If the function is a constant, its derivative is zero.

In our exercise, we see the use of the power rule to differentiate terms like \( 3t^5 \) which becomes \( 15t^4 \) after differentiation. Similarly, \( -t^2 \) differentiates to \( -2t \). Understanding these basic rules is essential for student success in calculus differentiation.

Understanding Polynomial Functions

Polynomials are expressions involving sums of powers of a variable multiplied by coefficients.
These functions are common in calculus and often show up in differentiation exercises.
A general polynomial function might look like \[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \] where each term is a power of \( x \) with a constant coefficient.

Differentiating polynomial functions involves applying the power rule to each term independently. Because polynomials are simply sums of terms, we can differentiate them term by term:

• Each power of \( t \) is reduced by one, and
• Each coefficient is multiplied by the original power of \( t \).

For example, differentiating the polynomial \( 3t^5 - t^2 \) results in \( 15t^4 - 2t \).
The process is straightforward as long as you apply the rules systematically to each part of the polynomial.