Derivative of a square root fraction. HELP

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Homework Help Overview

The problem involves finding the derivative of the function f(x) = 1/sqrt(2x). Participants are exploring various methods to compute the derivative, including applying the derivative formula and using exponent rules.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss different approaches to applying the derivative formula, including using limits and rewriting the function in exponent form. There are attempts to simplify the expression and questions about the appropriateness of the results obtained.

Discussion Status

The discussion is ongoing, with participants providing suggestions for rewriting the function and clarifying steps in the derivative process. Some guidance has been offered regarding the use of exponent notation, but there is no explicit consensus on the best approach yet.

Contextual Notes

Some participants express uncertainty about the manipulation of expressions involving radicals and the application of limit processes. There are indications of confusion regarding the correct application of the derivative rules and the handling of indeterminate forms.

curlybit89
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Homework Statement



What is f'(x) of f(x) 1/sqrt(2x)?

2. The attempt at a solution

In applying the problem to the derivative formula:

(1 / sqrt(2(x + h)) - 1 / sqrt(2x)) / h

I multiplied the problem by a special form of one but that only put the rationals on the bottom of the division. This looked too messy.

Alternatively, multiplying each side of the first division by it's denominator yielded the following:

(sqrt(2x) - sqrt(2x2h) / sqrt(2x^2 * 2x * 2h)) / h

I'm not sure how to proceed from here.

Another alternative I attempted was to use the exponent method which for me yielded:

original: 2x^-(1/2) = -(x)^-(3/2)

This answer did not seem appropriate.
 
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Think of another way to write the derivate. What is another way to write 1/sqrt(2x) ?

Thanks
Matt
 
Easiest solution in my opinion would be to just re-write it as [tex](2x)^{-1/2}[/tex] and simply take the derivative of that.

Don't forget it is f'(x)=n*[tex](u)^{n-1}[/tex] *u'
 
curlybit89 said:

Homework Statement



What is f'(x) of f(x) 1/sqrt(2x)?

2. The attempt at a solution

[snip]

Another alternative I attempted was to use the exponent method which for me yielded:

original: 2x^-(1/2) = -(x)^-(3/2)

This answer did not seem appropriate.

You are almost correct here.

[tex]\frac{1}{\sqrt{2x}} = (2x)^{-1/2}[/tex]

[tex]\frac{d}{dx}\left( (2x)^{-1/2} \right) = -\frac{1}{2}(\bold{2}x)^{-3/2}(2)[/tex]

Proceed from there...

--Elucidus
 
curlybit89 said:

Homework Statement



What is f'(x) of f(x) 1/sqrt(2x)?

2. The attempt at a solution

In applying the problem to the derivative formula:

(1 / sqrt(2(x + h)) - 1 / sqrt(2x)) / h

I multiplied the problem by a special form of one but that only put the rationals on the bottom of the division. This looked too messy.

Alternatively, multiplying each side of the first division by it's denominator yielded the following:

((sqrt(2x) - sqrt(2x + 2h)) / sqrt(2x^2 * 2x * 2h)) / h

You are missing a plus, and a pair of parentheses there.

When solving limit of the Indeterminate Form 0/0, you should try to factorize it. And factorizing the numerator, which contains radicals is impossible, right? So, one should think of a way to rationalize it. How about multiplying both numerator and denominator by

[tex]\sqrt{2x} + \sqrt{2x + 2h}[/tex]?
 

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