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Find the derivative of the following functions. See Example 4 of Section 3.1 for the derivative of \(\sqrt{x}\). $$g(t)=6 \sqrt{t}$$

Short Answer

Expert verified
Answer: The derivative of the function \(g(t) = 6\sqrt{t}\) is \(g'(t) = \frac{3}{\sqrt{t}}\).

Step by step solution

01

Rewrite the function with exponent notation

Rewrite the function \(g(t)=6\sqrt{t}\) using exponent notation. The square root of t can be written as \(t^{\frac{1}{2}}\). So the function becomes: $$g(t) = 6t^{\frac{1}{2}}$$
02

Apply the power rule

Apply the power rule of differentiation: \((x^n)' = nx^{n-1}\). Derive the function \(g(t) = 6t^{\frac{1}{2}}\) with respect to \(t\): $$(6t^{\frac{1}{2}})' = 6(\frac{1}{2}t^{\frac{1}{2}-1})$$
03

Simplify the expression

Simplify the expression to obtain the derivative of \(g(t)\): $$g'(t)= 6\left(\frac{1}{2}t^{-\frac{1}{2}}\right) = 3t^{-\frac{1}{2}}$$
04

Rewrite the derivative in the square root notation

Rewrite the derivative in the square root notation: $$g'(t) = \frac{3}{\sqrt{t}}$$ The derivative of the function \(g(t) = 6\sqrt{t}\) is \(g'(t) = \frac{3}{\sqrt{t}}\).

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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 a fundamental concept in calculus for finding derivatives of monomials, where the function is of the form f(x) = ax^n. It stipulates that the derivative of such a function is found by bringing down the exponent as a multiplier and then reducing the power by one. The generalized formula is \( (x^n)'=nx^{n-1} \).

Here's how to apply it step by step:
  • Identify the coefficient a and the power n in the function.
  • Multiply the coefficient by the power, which becomes the new coefficient.
  • Decrease the original power by one to find the new exponent.
  • Write the new function, which is the derivative of the original.
In the example of \( g(t)=6t^{\frac{1}{2}} \), applying the power rule gives us the derivative \( g'(t)=3t^{-\frac{1}{2}} \).

Understanding the power rule is crucial as it is one of the simplest and most used derivative rules in calculus, applicable to a wide range of problems involving powers of x.
Derivative of Square Root Functions
When we encounter functions that include square roots, we can utilize our knowledge of exponent notation to find derivatives more easily. A square root function can be written with an exponent of 1/2, turning it into a function applicable for the power rule. For a function like \( f(x) = \sqrt{x} \), using exponent notation, it becomes \( f(x) = x^{1/2} \).

To find its derivative, simply apply the power rule:
  • Rewrite the square root function using exponent notation \( x^{1/2} \) for \( \sqrt{x} \).
  • Apply the power rule, resulting in \( \frac{1}{2}x^{1/2-1} \) or \( \frac{1}{2}x^{-1/2} \).
  • Simplify further to get the final expression of the derivative.
Therefore, the derivative of the square root function \( f(x) = \sqrt{x} \) is \( f'(x) = \frac{1}{2\sqrt{x}} \) or in exponent notation \( \frac{1}{2}x^{-1/2} \). This method translates to more complex square root functions like \( 6\sqrt{t} \) by applying the same principles.
Exponent Notation
Exponent notation is a concise way to represent repeated multiplication of the same number. It is written as \( b^n \) where b is the base and n is the exponent, indicating how many times b is to be used as a factor.

In calculus, exponent notation is particularly useful because it simplifies the expression of roots and powers, making differentiation and integration procedures more straightforward.
  • The square root, cube root, and higher roots of a number x are expressed as \( x^{1/2}, x^{1/3}, x^{1/4}, \dots \), respectively.
  • Negative exponents indicate division by that number to a positive power, for example, \( x^{-n} = \frac{1}{x^n} \).
  • Fractional exponents simplify expressions involving roots and allow the application of rules like the power rule.
In the context of derivatives, translating functions into exponent notation not only lays the groundwork for the application of derivative rules but also leads to a neat expression of the result, which can be translated back into root or fractional form for final answers.

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Most popular questions from this chapter

Gone fishing When a reservoir is created by a new dam, 50 fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is \(P(t)=\frac{400,000}{50+7950 e^{-0.5 t}},\) where \(t\) is measured in years.c. How fast (in fish per year) is the population growing at \(t=0 ?\) At \(t=5 ?\) d. Graph \(P^{\prime}\) and use the graph to estimate the year in which the population is growing fastest.

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Find the following higher-order derivatives. $$\left.\frac{d^{3}}{d x^{3}}\left(x^{4.2}\right)\right|_{x=1}$$

Savings plan Beginning at age \(30,\) a self-employed plumber saves \(\$ 250\) per month in a retirement account until he reaches age \(65 .\) The account offers \(6 \%\) interest, compounded monthly. The balance in the account after \(t\) years is given by \(A(t)=50,000\left(1.005^{12 t}-1\right)\) a. Compute the balance in the account after \(5,15,25,\) and 35 years. What is the average rate of change in the value of the account over the intervals \([5,15],[15,25],\) and [25,35]\(?\) b. Suppose the plumber started saving at age 25 instead of age 30\. Find the balance at age 65 (after 40 years of investing). c. Use the derivative \(d A / d t\) to explain the surprising result in part (b) and to explain this advice: Start saving for retirement as early as possible.

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