Chapter 33: Problem 10
If \(f(t)=5 t+\frac{1}{\sqrt{t^{3}}}\) find \(f^{\prime}(t)\)
Short Answer
Expert verified
The derivative is \( f'(t) = 5 - \frac{3}{2t^{5/2}} \).
Step by step solution
01
Differentiate the Linear Term
The function given is \( f(t) = 5t + \frac{1}{\sqrt{t^3}} \). The first term is \( 5t \). The derivative of a linear function \( at \) with respect to \( t \) is simply the constant \( a \). So, for \( 5t \), the derivative is \( 5 \).
02
Rewrite the Second Term for Differentiation
The second term is \( \frac{1}{\sqrt{t^3}} \). This can be rewritten as \( t^{-3/2} \) by using the rule that \( \sqrt{t^3} = t^{3/2} \) and then applying the reciprocal rule to get \( t^{-3/2} \).
03
Differentiate the Rewritten Second Term
Using the power rule for differentiation, \( \frac{d}{dt} [t^n] = n \cdot t^{n-1} \), we differentiate \( t^{-3/2} \). Applying the power rule, we get the derivative: \( -\frac{3}{2}t^{-3/2-1} = -\frac{3}{2}t^{-5/2} \).
04
Combine the Derivatives
Now we combine the derivatives from Step 1 and Step 3. The derivative of the function \( f(t) \) is \( f'(t) = 5 + \left(-\frac{3}{2}t^{-5/2}\right) \). This simplifies to \( f'(t) = 5 - \frac{3}{2t^{5/2}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus. It is used to find the rate at which a function is changing at any given point. Imagine a car traveling along a road; differentiation helps us find out how fast the car is moving at any specific instant. In mathematical terms, it provides us with the derivative of a function. This process breaks down complex functions into simpler ones, enabling us to analyze their behavior more easily. For this exercise, remember the function involves both a linear and a non-linear term, and each part requires handling them separately during differentiation.
Power Rule
The power rule is one of the most straightforward techniques for differentiation. It allows us to find the derivative of polynomial expressions easily.
This rule states that if you have a function expressed as a power, like \(t^n\), its derivative is \(n \cdot t^{n-1}\). This means you bring down the exponent as a coefficient and subtract one from the original exponent.
This rule states that if you have a function expressed as a power, like \(t^n\), its derivative is \(n \cdot t^{n-1}\). This means you bring down the exponent as a coefficient and subtract one from the original exponent.
- For example, the derivative of \(t^3\) is \(3t^2\).
- The derivative of \(t^{-4}\) is \(-4t^{-5}\).
Derivative of Functions
The derivative of a function provides critical information about the function's behavior—such as its increasing or decreasing behavior at certain intervals. For the given function, \(f(t) = 5t + \frac{1}{\sqrt{t^3}}\), we have to find its derivative \(f'(t)\).
Here, each part of the function contributes to the total derivative:
Here, each part of the function contributes to the total derivative:
- For \(5t\), the derivative is a simple constant due to the power of 1.
- For the complex term \(\frac{1}{\sqrt{t^3}}\), rewriting it helps simplify differentiation through the power rule.
Rewriting Expressions
Rewriting expressions in a differentiable form is a useful strategy in calculus because it simplifies the process of finding derivatives. In this exercise, the term \(\frac{1}{\sqrt{t^3}}\) was initially challenging to differentiate. By rewriting it as \(t^{-3/2}\), it becomes more manageable.
- Using rules of exponents, \(\sqrt{t^3}\) becomes \(t^{3/2}\).
- Applying the reciprocal, \(\frac{1}{t^{3/2}}\) can be rewritten as \(t^{-3/2}\).