Derivative of a square root fraction. HELP

In summary: This would result in((sqrt(2x) - sqrt(2x + 2h)) / (sqrt(2x) + sqrt(2x + 2h))) * ((sqrt(2x) + sqrt(2x + 2h)) / sqrt(2x^2 * 2x * 2h)) / h= (2x - 2x - 2h) / (2x^2 * 2x * 2h * h * (sqrt(2x) + sqrt(2x + 2h)))= (-2h) / (2x^2 * 2x * 2h * h * (sqrt(2x) +
  • #1
curlybit89
2
0

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.
 
Physics news on Phys.org
  • #2
Think of another way to write the derivate. What is another way to write 1/sqrt(2x) ?

Thanks
Matt
 
  • #3
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'
 
  • #4
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
 
  • #5
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]?
 

What is the derivative of a square root fraction?

The derivative of a square root fraction is calculated by applying the quotient rule of differentiation, which states that the derivative of a fraction is equal to the denominator multiplied by the derivative of the numerator, minus the numerator multiplied by the derivative of the denominator, all divided by the square of the denominator.

How do you find the derivative of a square root fraction?

To find the derivative of a square root fraction, first rewrite it as a fraction with a power of -1/2. Then, apply the quotient rule of differentiation to find the derivative. Finally, rewrite the answer in its original form.

Can the derivative of a square root fraction be simplified?

Yes, the derivative of a square root fraction can be simplified by using algebraic manipulation techniques such as factoring, simplifying, and combining like terms. It is important to simplify the answer in order to have a clear and concise understanding of the derivative.

What is the purpose of finding the derivative of a square root fraction?

The purpose of finding the derivative of a square root fraction is to determine the rate of change, or slope, of the function at a specific point. This is useful in many areas of science, such as physics, engineering, and economics, where understanding the rate of change is essential.

Are there any special cases when finding the derivative of a square root fraction?

Yes, there are special cases when finding the derivative of a square root fraction. One special case is when the denominator is a constant, in which case the derivative simplifies to the negative constant multiplied by the square root of the numerator. Another special case is when the numerator is a constant, in which case the derivative simplifies to the constant divided by the square root of the denominator.

Similar threads

Replies
7
Views
518
  • Calculus and Beyond Homework Help
Replies
4
Views
918
  • Calculus and Beyond Homework Help
Replies
4
Views
559
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
818
  • Calculus and Beyond Homework Help
Replies
20
Views
453
  • Calculus
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top