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Find \(y^{\prime}\). $$ y=\frac{\ln \sqrt{2 x}}{x} $$

Short Answer

Expert verified
The derivative of the function \(y=\frac{\ln \sqrt{2x}}{x}\) is \(y^{\prime}=\frac{\sqrt{(x/2)}}{x^{2}} - \frac{\ln \sqrt{2x}}{x^{2}})\

Step by step solution

01

Identify the Top Function and Bottom Function

First, recognize \(y=\frac{\ln \sqrt{2x}}{x}\) as a quotient of two functions, namely \(f(x)=\ln \sqrt{2x}\) and \(g(x)=x\). In finding the derivative of this quotient, the quotient rule will be needed.
02

Apply the Quotient Rule

The quotient rule states that the derivative of the quotient of two functions is given by \((f(x) / g(x))' = (f'(x)g(x) - f(x)g'(x)) / (g(x))^{2}\). So, apply the rule to the function, but leave \(f'(x)\) and \(g'(x)\) as placeholders for now.
03

Find the Derivative of the Top Function

To find the derivative of \(f(x)=\ln \sqrt{2x}\), the chain rule must be used. The chain rule states that the derivative of a composite function is the product of the derivative of the outside function and the derivative of the inside function. Here, the outside function can be seen as \(u=\ln(x)\) and the inside function as \(v=\sqrt{2x}\). Differentiating these provides the derivative of the top function.
04

Calculate the Derivative of the Bottom Function

The derivative of \(g(x)=x\) is quite simple. Applying the power rule where the derivative of \(x^n\) is \(nx^{n-1}\), with \(n=1\) in this case, yields the derivative of the bottom function to be 1.
05

Substitute Back in the Function Derivatives

Substitute the derivative of the top function, \(f'(x)\), and the derivative of the bottom function, \(g'(x)\), into the function with the placeholders mentioned in Step 2. This provides the derivative of the original function.

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Key Concepts

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

Quotient Rule
The quotient rule is a powerful technique in calculus used to find the derivative of a function that is expressed as a quotient, or a division, of two other functions. If you have a function that can be written as one function divided by another, like \(y = \frac{u(x)}{v(x)}\), the quotient rule is your go-to tool.

Applying the quotient rule, the derivative is found using the formula \(y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). To put it simply, you take the derivative of the top function \(u(x)\) and multiply it by the bottom function \(v(x)\), then subtract the product of the top function and the derivative of the bottom function \(v'(x)\). Finally, you divide the whole thing by the bottom function squared.

For instance, if you're differentiating \(y = \frac{\ln \sqrt{2x}}{x}\), you'd start by labeling \(\ln \sqrt{2x}\) as \(u(x)\) and \(x\) as \(v(x)\), and then apply the steps of the quotient rule. It's essential to proceed methodically to avoid errors, as both finding the derivatives and the algebra involved can become complex.
Chain Rule
The chain rule is a critical principle when it comes to differentiating composite functions—functions made up of two or more functions that are combined in some way. When you have a function contained within another function, the chain rule allows you to differentiate it piece by piece.

Formally, if you have a composite function \(y = h(g(x))\), the derivative would be \(y' = h'(g(x)) \cdot g'(x)\). Here, you're essentially differentiating the outer function \(h\) with respect to the inner function \(g\), and then multiplying it by the derivative of the inner function \(g\) with respect to \(x\).

For example, when differentiating \(\ln \sqrt{2x}\) as in the exercise, you're dealing with the logarithm of a square root, meaning you must differentiate the logarithm function while also accounting for the square root inside of it. Carefully applying the chain rule lets you break down this more complex derivative into manageable parts.
Natural Logarithm Differentiation
Differentiating natural logarithms involves a unique rule. The derivative of the natural logarithm of \(x\), represented as \(\ln(x)\), is \(1/x\). The process becomes more intricate when the logarithm has a composite function as its argument, such as \(\ln \sqrt{2x}\).

In such cases, you must again invoke the chain rule. For \(\ln \sqrt{2x}\), you can't merely apply the simple rule of \(1/x\), because \(\sqrt{2x}\) isn't just \(x\); it's another function of \(x\). Instead, find the derivative of the inside function \(\sqrt{2x}\) and multiply it by the derivative of the outer function, which in this case is \(\ln(u)\) where \(u = \sqrt{2x}\).

This may sound daunting, but with practice, the method becomes more natural. Remember, the derivative of the natural logarithm is not confined within its simple rule but must be adapted when dealing with more complex functions.
Power Rule
The power rule is a fundamental concept that makes differentiating functions of the form \(x^n\) quite straightforward. According to this rule, if you have a variable \(x\) raised to a power \(n\), its derivative is \(nx^{n-1}\). This means you multiply the original exponent by the variable, and then subtract one from the exponent.

In the context of our exercise, you'd apply the power rule to differentiate the bottom function \(g(x) = x\). The exponent here is 1, so according to the power rule, the derivative \(g'(x)\) of \(x\) is \(1 \cdot x^{1-1} = x^0 = 1\).

The power rule simplifies the differentiation of polynomial functions significantly, making it easier to compute derivatives and hence making calculus much more approachable for a wide range of problems.

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Most popular questions from this chapter

In early summer the fly population in Maine grows exponentially. The population at any time \(t\) can be given by \(P(t)=P_{0} e^{k t}\) for some constant \(k\), where \(t\) is measured in days. Suppose that at some date, which we will designate as \(t=0\), there are 200 flies. Thirty days later there are 900 flies. (a) Find the constant \(k\). (b) The mosquito population is also growing exponentially. At time \(t=0\) there are 100 mosquitoes, and the mosquito population doubles every 10 days. Write a function \(M(t)\) that gives the number of mosquitoes at time \(t\). (c) When will the number of flies and the number of mosquitoes be equal? (d) Find \(P^{\prime}(t)\). (e) Find \(M^{\prime}(t)\). (f) Find the rate at which each of the populations is growing when the populations are the same size. Which is growing more rapidly?

Differentiate the given function. $$ f(x)=x^{2} \ln \left(x \sqrt{\frac{61}{2 x}}\right) $$

Let \(f(x)=\ln x-x\). (a) What is the domain of this function? (b) Find all the critical points of \(f\). (The critical points must be in the domain of \(f .\) ) (c) By looking at the sign of \(f^{\prime}\), find all local maxima and minima. Give both the \(x\) and \(y\) -coordinates of the extrema. (d) Find \(f^{\prime \prime}\). Where is \(f\) concave up and where is \(f\) concave down? (e) Sketch the graph of \(\ln x-x\) without using a calculator (except possibly to check your work).

Using what you know about the graph of \(\ln x\), sketch the graphs of the following. (a) \(y=|\ln x|\) (b) \(y=\ln (|x|)\) (c) \(y=|\ln (|x|)|\) (d) For parts (a), (b), and (c), locate all critical points and identify all local maxima and minima.

Suppose you put \(\$ 500\) in a bank account and your balance grows exponentially according to the equation $$ M=500 e^{0.08 t}, $$ where \(M=M(t)=\) the amount of money in the account at time \(t\). (a) Write the growth equation for the amount of money in the account in the form \(M=500 A^{t}\) (b) What is the annual growth rate of the money in the account? (Banks refer to this as the effective annual yield.) Please give your answer to the nearest tenth of a percent. (c) What is the instantaneous rate of change of money with respect to time? (d) When will you have enough money to buy a round-the-world plane ticket costing \(\$ 1599 ?\)

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