Finding Derivatives Using Differentiation Rules A Comprehensive 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 numerous applications in various fields such as physics, engineering, economics, and computer science. This article provides a detailed explanation of how to find derivatives using the rules of differentiation, with a focus on simplifying the results. We will explore two specific examples, demonstrating the application of the chain rule, quotient rule, and power rule.

(i) Differentiating $f(x) = \left(\frac{x-1}{2x+1}\right)^5$

In this section, we tackle the problem of finding the derivative of the function $f(x) = \left(\frac{x-1}{2x+1}\right)^5$. This function involves a composite structure, requiring the application of multiple differentiation rules. Specifically, we will utilize the chain rule in conjunction with the quotient rule and the power rule. The chain rule is essential for differentiating composite functions, where one function is nested inside another. The quotient rule is necessary for handling the fraction within the parentheses, and the power rule applies to the outer exponent. To effectively navigate this differentiation, we'll break down the process into manageable steps, ensuring clarity and accuracy in our calculations. Understanding each step is crucial for mastering differentiation techniques and applying them to more complex problems.

Step-by-Step Differentiation

1. Applying the Chain Rule:

   • The chain rule states that if we have a composite function $f(g(x))$, its derivative is $f'(g(x)) \\cdot g'(x)$. In our case, we can consider the outer function as $u^5$ and the inner function as $u = \frac{x-1}{2x+1}$.

   • Therefore, the derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$, can be written as:

     $f'(x) = 5\left(\frac{x-1}{2x+1}\right)^4 \\cdot \frac{d}{dx}\left(\frac{x-1}{2x+1}\right)$

2. Applying the Quotient Rule:

   • Now we need to find the derivative of the inner function $\frac{x-1}{2x+1}$. The quotient rule states that if we have a function $h(x) = \frac{p(x)}{q(x)}$, its derivative is $h'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{[q(x)]^2}$.

   • Here, $p(x) = x-1$ and $q(x) = 2x+1$. Thus, $p'(x) = 1$ and $q'(x) = 2$.

   • Applying the quotient rule, we get:

     $\frac{d}{dx}\left(\frac{x-1}{2x+1}\right) = \frac{1 \\cdot (2x+1) - (x-1) \\cdot 2}{(2x+1)^2}$

3. Simplifying the Derivative of the Inner Function:

   • Let's simplify the expression obtained from the quotient rule:

     $\frac{1 \\cdot (2x+1) - (x-1) \\cdot 2}{(2x+1)^2} = \frac{2x+1 - 2x + 2}{(2x+1)^2} = \frac{3}{(2x+1)^2}$

4. Substituting Back into the Chain Rule Expression:

   • Now, we substitute the simplified derivative of the inner function back into the chain rule expression:

     $f'(x) = 5\left(\frac{x-1}{2x+1}\right)^4 \\cdot \frac{3}{(2x+1)^2}$

5. Further Simplification:

   • To further simplify, we can combine the terms:

     $f'(x) = \frac{15(x-1)^4}{(2x+1)^4 \\cdot (2x+1)^2} = \frac{15(x-1)^4}{(2x+1)^6}$

Final Result

The derivative of the function $f(x) = \left(\frac{x-1}{2x+1}\right)^5$ is:

$\boxed{f'(x) = \frac{15(x-1)^4}{(2x+1)^6}}$

This result showcases the application of the chain rule and the quotient rule in finding derivatives. The step-by-step approach allows for clear understanding and minimizes the chances of error. Mastering these rules is fundamental for tackling more complex differentiation problems in calculus.

(ii) Differentiating $g(x) = \sin^2(\sqrt[3]{x})$

In this section, we delve into finding the derivative of the function $g(x) = \sin^2(\sqrt[3]{x})$. This function, similar to the previous example, requires the application of the chain rule due to its composite nature. However, in this case, we have a triple composite function, meaning we have three functions nested within each other. These functions are the squaring function, the sine function, and the cube root function. To differentiate such a function effectively, we need to apply the chain rule multiple times, peeling away the layers of composition one by one. This process involves carefully identifying the inner and outer functions at each step and applying the chain rule formula iteratively. Understanding how to handle multiple compositions is crucial for differentiating complex functions and is a fundamental skill in calculus. This section will provide a detailed breakdown of each step, ensuring a clear understanding of the differentiation process.

Step-by-Step Differentiation

1. Applying the Chain Rule (First Layer):

   • We can rewrite $g(x)$ as $(\sin(\sqrt[3]{x}))^2$. Applying the chain rule, we treat the outer function as $u^2$ and the inner function as $u = \sin(\sqrt[3]{x})$.

   • The derivative of $g(x)$ with respect to $x$, denoted as $g'(x)$, can be written as:

     $g'(x) = 2\sin(\sqrt[3]{x}) \\cdot \frac{d}{dx}(\sin(\sqrt[3]{x}))$

2. Applying the Chain Rule (Second Layer):

   • Now we need to find the derivative of $\sin(\sqrt[3]{x})$. Again, we apply the chain rule, treating the outer function as $\sin(v)$ and the inner function as $v = \sqrt[3]{x}$.

   • The derivative of $\sin(\sqrt[3]{x})$ with respect to $x$ is:

     $\frac{d}{dx}(\sin(\sqrt[3]{x})) = \cos(\sqrt[3]{x}) \\cdot \frac{d}{dx}(\sqrt[3]{x})$

3. Differentiating the Innermost Function:

   • We now need to find the derivative of $\sqrt[3]{x}$, which can be written as $x^{\frac{1}{3}}$. Applying the power rule, we get:

     $\frac{d}{dx}(x^{\frac{1}{3}}) = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$

4. Substituting Back into the Chain Rule Expressions:

   • We now substitute the derivatives back into the previous expressions. First, we substitute the derivative of $\sqrt[3]{x}$:

     $\frac{d}{dx}(\sin(\sqrt[3]{x})) = \cos(\sqrt[3]{x}) \\cdot \frac{1}{3x^{\frac{2}{3}}}$

   • Then, we substitute this result back into the expression for $g'(x)$:

     $g'(x) = 2\sin(\sqrt[3]{x}) \\cdot \cos(\sqrt[3]{x}) \\cdot \frac{1}{3x^{\frac{2}{3}}}$

5. Simplifying the Derivative:

   • To simplify, we can combine the terms:

     $g'(x) = \frac{2\sin(\sqrt[3]{x})\cos(\sqrt[3]{x})}{3x^{\frac{2}{3}}}$

   • Using the trigonometric identity $2\sin(a)\cos(a) = \sin(2a)$, we can further simplify the numerator:

     $g'(x) = \frac{\sin(2\sqrt[3]{x})}{3x^{\frac{2}{3}}}$

Final Result

The derivative of the function $g(x) = \sin^2(\sqrt[3]{x})$ is:

$\boxed{g'(x) = \frac{\sin(2\sqrt[3]{x})}{3x^{\frac{2}{3}}}}$

This example demonstrates the power of the chain rule in handling triple composite functions. By systematically applying the chain rule and simplifying the results, we can find the derivatives of complex functions. Understanding the underlying principles and practicing these techniques are essential for mastering differential calculus.

Conclusion

Finding the derivatives of functions is a crucial skill in calculus. Through the examples presented, we have explored the application of the chain rule, quotient rule, and power rule. Mastering these rules and techniques allows for the effective differentiation of a wide range of functions. The step-by-step approach, as demonstrated in this article, is crucial for ensuring accuracy and clarity in the differentiation process. By breaking down complex problems into smaller, manageable steps, we can confidently tackle even the most challenging differentiation tasks. Continuous practice and a deep understanding of these fundamental rules will pave the way for advanced applications of calculus in various scientific and engineering disciplines.