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Find the derivative. Assume \(a, b, c, k\) are constants. $$P=a+b \sqrt{t}$$

Short Answer

Expert verified
\(\frac{dP}{dt} = \frac{b}{2\sqrt{t}}\)

Step by step solution

01

Identify the Function to Differentiate

The function given is a polynomial form with a square root term: \[ P(t) = a + b \sqrt{t} \]Here, \(a\) and \(b\) are constants, and \(\sqrt{t}\) is a function of \(t\).
02

Recall the Power Rule for Derivatives

The power rule states that for any function \(t^n\), the derivative \(\frac{d}{dt}t^n = n \cdot t^{n-1} \). We will use this rule for \(\sqrt{t}\), noting that \(\sqrt{t} = t^{1/2}\).
03

Differentiate Each Term Separately

1. The derivative of the constant term \(a\) with respect to \(t\) is 0.2. For the term \(b \sqrt{t} = b t^{1/2}\): - Apply the power rule: \(\frac{d}{dt} t^{1/2} = \frac{1}{2} t^{-1/2}\). - The derivative of \(b t^{1/2}\) is \(b \cdot \frac{1}{2} t^{-1/2} = \frac{b}{2} t^{-1/2}\).Thus, \[ \frac{dP}{dt} = 0 + \frac{b}{2} t^{-1/2} \].
04

Write the Final Expression in Terms of \(t\)

The expression \(t^{-1/2}\) can be rewritten using the square root: \(t^{-1/2} = \frac{1}{\sqrt{t}}\). Hence, the derivative in terms of \(t\) is:\[ \frac{dP}{dt} = \frac{b}{2 \sqrt{t}} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule in Derivatives
The power rule is a fundamental concept in calculus that helps simplify the process of finding derivatives. It applies to functions where a variable is raised to a power, such as \( t^n \). The rule states that the derivative of \( t^n \) with respect to \( t \) is given by \( \frac{d}{dt}t^n = n \cdot t^{n-1} \). This means you multiply the original power by the variable, then reduce the power by one. For example, the derivative of \( t^{3} \) would be \( 3t^{2} \). Similarly, when you encounter square roots in the function, you can rewrite them as fractional powers, like \( \sqrt{t} = t^{1/2} \), and apply the power rule just the same. This method makes differentiating complex expressions more straightforward by breaking them down into manageable parts. The power rule is especially useful because of its simplicity, making it an essential tool in both basic and advanced calculus problems.
Square Root Differentiation
Differentiating square roots can seem tricky at first, but they become clearer when you remember they’re just exponents in disguise. Any square root can be expressed as a power of \( 1/2 \). For instance, \( \sqrt{t} \) becomes \( t^{1/2} \). This transformation is key because it allows you to apply the power rule. When you differentiate \( t^{1/2} \) using the power rule, you bring down the power (\( 1/2 \)) and subtract one from it. This leaves you with \( \frac{1}{2}t^{-1/2} \). Essentially, differentiating square roots is simplified to a two-step process: writing the root as a power and applying the power rule. This method not only clarifies the process but also retains simplicity and consistency across various functions involving roots. It's a handy technique in calculus that ensures you can handle even more challenging differentiation tasks with confidence.
Constant Rule in Derivatives
In calculus, constants have a special set of rules for differentiation that greatly simplify calculations. The constant rule states that the derivative of any constant is zero. This is because constants do not change, and derivatives measure how a quantity changes. If it doesn’t change, its rate of change or derivative is obviously zero. For example, if you have a function like \( P(t) = 7 \), its derivative \( \frac{d}{dt} 7 \) is simply 0. Similarly, in our original function \( P(t) = a + b \sqrt{t} \), the term \( a \) is a constant, which means its derivative with respect to \( t \) is also 0. Understanding this rule enhances our ability to dissect functions into simpler parts, focusing only on the variable-dependent portions when computing derivatives. This rule is indispensable in both simple and complex calculus tasks and is part of the foundational concepts you'll use frequently.

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