Derivative of a square root fraction. HELP

curlybit89
Messages
2
Reaction score
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
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 (2x)^{-1/2} and simply take the derivative of that.

Don't forget it is f'(x)=n*(u)^{n-1} *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.

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

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

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

\sqrt{2x} + \sqrt{2x + 2h}?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top