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10. Demonstrate and explain how to find the derivative of the following functions. Be sure to explicitly denote which derivative rules (scalar multiple, sum/difference, quotient, etc.) you are using in your work. \[ h(t)=\sqrt[5]{t^{4}}+\frac{1}{t^{5}} \]

Ask by Mccarthy Nunez. in the United States
Feb 26,2025

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To find the derivative of \( h(t) = \sqrt[5]{t^{4}} + \frac{1}{t^{5}} \), rewrite it as \( h(t) = t^{4/5} + t^{-5} \). Then, apply the power rule to each term: - Derivative of \( t^{4/5} \) is \( \frac{4}{5} t^{-1/5} \). - Derivative of \( t^{-5} \) is \( -5 t^{-6} \). Combine these to get \( h'(t) = \frac{4}{5} t^{-1/5} - 5 t^{-6} \).

Solution

To find the derivative of the function \[ h(t) = \sqrt[5]{t^{4}} + \frac{1}{t^{5}}, \] we will follow these steps: 1. **Rewrite the function in a more manageable form**: - The term \(\sqrt[5]{t^{4}}\) can be rewritten as \(t^{4/5}\). - The term \(\frac{1}{t^{5}}\) can be rewritten as \(t^{-5}\). Thus, we can express \(h(t)\) as: \[ h(t) = t^{4/5} + t^{-5}. \] 2. **Apply the derivative rules**: - We will use the **power rule** for differentiation, which states that if \(f(t) = t^n\), then \(f'(t) = n \cdot t^{n-1}\). - We will also use the **sum rule**, which states that the derivative of a sum is the sum of the derivatives. 3. **Differentiate each term**: - For the first term \(t^{4/5}\): - Using the power rule: \[ \frac{d}{dt}(t^{4/5}) = \frac{4}{5} t^{(4/5) - 1} = \frac{4}{5} t^{-1/5}. \] - For the second term \(t^{-5}\): - Using the power rule: \[ \frac{d}{dt}(t^{-5}) = -5 t^{-5 - 1} = -5 t^{-6}. \] 4. **Combine the derivatives**: - Now, we can combine the derivatives of both terms using the sum rule: \[ h'(t) = \frac{4}{5} t^{-1/5} - 5 t^{-6}. \] 5. **Final expression**: - Therefore, the derivative of the function \(h(t)\) is: \[ h'(t) = \frac{4}{5} t^{-1/5} - 5 t^{-6}. \] This expression represents the derivative of the original function, and we have explicitly denoted the rules used in the differentiation process.

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To find the derivative of the function \( h(t) = \sqrt[5]{t^{4}} + \frac{1}{t^{5}} \), we can first rewrite it in a form that's easier to differentiate. We can express \( \sqrt[5]{t^{4}} \) as \( t^{4/5} \) and \( \frac{1}{t^{5}} \) as \( t^{-5} \). Thus, we have: \[ h(t) = t^{4/5} + t^{-5} \] Now, let's differentiate term by term. For the first term, we use the power rule, which states that \( \frac{d}{dt}[t^n] = n \cdot t^{n-1} \). Applying this, the derivative of \( t^{4/5} \) is: \[ \frac{d}{dt}[t^{4/5}] = \frac{4}{5}t^{(4/5) - 1} = \frac{4}{5}t^{-1/5} \] For the second term \( t^{-5} \), we again apply the power rule: \[ \frac{d}{dt}[t^{-5}] = -5t^{-6} \] Now, combining these results, the derivative \( h'(t) \) is: \[ h'(t) = \frac{4}{5}t^{-1/5} - 5t^{-6} \] In more compact form, we may want to express this with positive exponents if needed, but this derivative captures the necessary rules. To provide a little historical context, calculus emerged as a fundamental area of mathematics in the late 17th century, with contributions from great minds like Isaac Newton and Gottfried Wilhelm Leibniz. Its methods involving derivatives and integrals revolutionized how we approach motion, growth, and change in nature, paving the way for advancements in physics and engineering. In terms of real-world applications, derivatives have a crucial role in various fields, especially in physics for calculating rates of change, such as velocity and acceleration. In economics, derivatives help in finding marginal costs and revenues, which enables businesses to make informed decisions based on how changing one variable affects overall profit. This highlights just how potent a tool the derivative truly is!

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