Derivative Of G(t) = 5/(6t^5) A Step-by-Step Guide
In calculus, finding the derivative of a function is a fundamental operation. The derivative represents the instantaneous rate of change of a function, and it has wide applications in various fields, including physics, engineering, economics, and computer science. In this article, we will explore how to find the derivative of the function g(t) = 5/(6t^5) using differentiation rules. We'll break down the process step-by-step, making it easy to understand even if you're new to calculus.
Understanding the Function
Before we dive into the differentiation process, let's take a closer look at the function we're dealing with:
g(t) = 5/(6t^5)
This function can be rewritten to make it easier to differentiate. By bringing the t^5 term from the denominator to the numerator, we get:
g(t) = (5/6) * t^(-5)
Rewriting the function in this form allows us to readily apply the power rule, a crucial differentiation rule that we'll discuss in detail later.
Essential Differentiation Rules
To find the derivative of g(t), we'll primarily use the power rule and the constant multiple rule. Let's briefly discuss these rules:
1. The Power Rule
The power rule is one of the most fundamental differentiation rules. It states that if you have a function of the form f(x) = x^n, where n is a constant, then the derivative of f(x) with respect to x is:
f'(x) = n * x^(n-1)
In simpler terms, the power rule says that you multiply the original exponent (n) by the variable (x) raised to the power of (n-1). This rule is widely applicable when dealing with polynomial terms and functions involving powers of a variable.
2. The Constant Multiple Rule
The constant multiple rule states that if you have a function of the form f(x) = c * g(x), where c is a constant, then the derivative of f(x) with respect to x is:
f'(x) = c * g'(x)
This rule essentially says that you can pull a constant factor out of the differentiation process. It simplifies the process of finding derivatives when a function is multiplied by a constant.
Applying the Differentiation Rules to g(t)
Now that we've discussed the essential differentiation rules, let's apply them to find the derivative of g(t) = (5/6) * t^(-5). We'll use both the constant multiple rule and the power rule.
Step 1: Apply the Constant Multiple Rule
First, we apply the constant multiple rule, which allows us to factor out the constant (5/6):
g'(t) = (5/6) * d/dt [t^(-5)]
This step simplifies the problem by isolating the term that needs differentiation.
Step 2: Apply the Power Rule
Next, we apply the power rule to differentiate t^(-5). According to the power rule, we multiply the exponent (-5) by t raised to the power of (-5 - 1):
d/dt [t^(-5)] = -5 * t^(-6)
Step 3: Combine the Results
Now, we substitute the result from Step 2 back into the equation from Step 1:
g'(t) = (5/6) * (-5 * t^(-6))
Step 4: Simplify the Expression
Finally, we simplify the expression by multiplying the constants and rewriting the term with a negative exponent:
g'(t) = -25/6 * t^(-6)
To eliminate the negative exponent, we can rewrite t^(-6) as 1/t^6:
g'(t) = -25 / (6t^6)
Therefore, the derivative of g(t) = 5/(6t^5) is g'(t) = -25 / (6t^6).
Alternative Method: Quotient Rule
While we successfully found the derivative using the power rule and constant multiple rule, there's another method we could have used: the quotient rule. The quotient rule is particularly useful when dealing with functions that are expressed as a ratio of two functions. Let's explore how we can apply the quotient rule to find the derivative of g(t) = 5/(6t^5).
Understanding the Quotient Rule
The quotient rule states that if you have a function of the form f(x) = u(x) / v(x), where u(x) and v(x) are differentiable functions, then the derivative of f(x) with respect to x is:
f'(x) = [u'(x) * v(x) - u(x) * v'(x)] / [v(x)]^2
In this rule:
- u'(x) is the derivative of the numerator function u(x).
- v'(x) is the derivative of the denominator function v(x).
Applying the Quotient Rule to g(t)
To apply the quotient rule to g(t) = 5/(6t^5), we first identify the numerator function u(t) and the denominator function v(t):
- u(t) = 5
- v(t) = 6t^5
Next, we find the derivatives of u(t) and v(t):
- u'(t) = d/dt [5] = 0 (since the derivative of a constant is zero)
- v'(t) = d/dt [6t^5] = 6 * 5 * t^(5-1) = 30t^4 (using the power rule)
Now, we substitute these values into the quotient rule formula:
g'(t) = [u'(t) * v(t) - u(t) * v'(t)] / [v(t)]^2 g'(t) = [0 * (6t^5) - 5 * (30t^4)] / [(6t5)]2
Simplifying the Expression
Next, we simplify the expression:
g'(t) = [-150t^4] / [36t^10]
We can further simplify by dividing both the numerator and the denominator by their greatest common factor, which is 6t^4:
g'(t) = -25 / (6t^6)
As we can see, the result obtained using the quotient rule is the same as the result we obtained using the power rule and constant multiple rule. This demonstrates that different differentiation rules can often be used to arrive at the same answer.
Conclusion
In this article, we successfully found the derivative of the function g(t) = 5/(6t^5) using differentiation rules. We explored the power rule and the constant multiple rule, which allowed us to efficiently find the derivative by rewriting the function and applying the rules step-by-step. Additionally, we demonstrated an alternative method using the quotient rule, which further solidified our understanding of differentiation techniques. Mastering differentiation rules is crucial for calculus, and it opens doors to solving a wide range of problems in mathematics and related fields. By practicing and understanding these rules, you'll be well-equipped to tackle more complex calculus problems.
Understanding how to find derivatives is not just an academic exercise; it's a powerful tool that can be applied in real-world scenarios. For instance, in physics, derivatives are used to calculate velocity and acceleration. In economics, they are used to determine marginal cost and marginal revenue. By grasping the concepts and techniques discussed in this article, you'll gain a valuable skill that can be used across various disciplines.
Keep practicing and exploring different functions to further enhance your differentiation skills. Calculus is a fascinating and powerful field, and mastering the basics will set you on a path to deeper understanding and application.
