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Chapter 3: Problem 16

Find the derivative of the following functions. See Example 4 of Section 1 for the derivative of \(\sqrt{x}\). \(g(t)=6 \sqrt{t}\)

Short Answer

Expert verified
Question: Find the derivative of the function \(g(t)=6\sqrt{t}\). Answer: The derivative of the function \(g(t)=6\sqrt{t}\) is \(g'(t)= 3t^{-1/2}\).

Step by step solution

01

Rewrite the function

Rewrite the given function \(g(t)=6\sqrt{t}\) in terms of a power function using the exponent 1/2 instead of the square root. That is: \(g(t)=6t^{1/2}\)
02

Apply the power rule for derivatives

Now apply the power rule \((t^n)'=nt^{n-1}\) to the function \(g(t)=6t^{1/2}\). Apply the rule to the \(t^{1/2}\) term and multiply by the constant 6: \( g'(t) = 6 \cdot (\frac{1}{2})t^{\frac{1}{2} - 1}\)
03

Simplify the expression

Simplify the expression obtained in step 2: \(g'(t)= 3t^{-\frac{1}{2}}\)
04

Express the final answer

The derivative of the given function \(g(t)=6\sqrt{t}\) is: \(g'(t)= 3t^{-1/2}\)

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Key Concepts

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

Power Rule
The power rule is an essential tool in calculus for taking derivatives, especially when dealing with power functions. To use the power rule, you follow a simple formula:
  • If you have a function of the form \(t^n\), then its derivative is \(nt^{n-1}\).
  • This means you multiply by the exponent and then reduce the exponent by one.
In the example given, the function was initially \(g(t)=6\sqrt{t}\). To apply the power rule, it is crucial to express the square root in power notation. By rewriting the square root \(\sqrt{t}\) as \(t^{1/2}\), we can easily apply the rule. The coefficient 6 remains multiplied with the result of the derivative. Thus, \[g'(t) = 6 \cdot (\frac{1}{2})t^{\frac{1}{2} - 1}\]makes it straightforward to calculate, as you multiply the original exponent by the coefficient and subtract one from the exponent.
Function Notation
In calculus, function notation is a method to denote and manage functions clearly, especially when discussing derivatives. Instead of writing equations with 鈥測鈥� or other variables, we use function notation such as \(g(t)\) to better understand its behavior.
  • The notation makes it clear that \(g\) is a function of \(t\).
  • This format is beneficial when taking derivatives multiple times and communicating the function's dynamics.
In our exercise, the function \(g(t)=6\sqrt{t}\) was transformed for differentiation purposes. It then becomes \(g'(t)\) once differentiated, indicating clearly that this notation refers to the derivative of \(g(t)\). Using the function notation, especially during calculus operations, helps distinguish between functions, their derivatives, and other transformations, maintaining clarity.
Simplifying Expressions
Simplifying expressions is a vital step in calculus as it helps in finalizing the derivatives and making them easier to interpret. When you simplify, you aim to present the expression in its cleanest form.
  • After applying the power rule, always ensure any constants are properly multiplied.
  • Adjust exponents and combine like terms if necessary.
In our example, after applying the power rule, the expression was \(g'(t) = 6 \cdot \frac{1}{2} t^{\frac{1}{2} - 1}\). Through simplifying, we adjust the coefficient: \(6 \times \frac{1}{2} = 3\), and simplify the exponent by performing the subtraction: \(\frac{1}{2} - 1 = -\frac{1}{2}\). This simplification gives us the cleaner and final answer: \(g'(t) = 3t^{-1/2}\). By simplifying the expression, the result becomes easier to understand and use in further calculations.

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Most popular questions from this chapter

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=\tan x ;(1, \pi / 4)$$

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