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y = \frac{\log_{e} x}{x} Letting u = \log_{e} x and v = x: u' = \frac{1}{x} v' = 1 \frac{dy}{dx} = \frac{vu' - uv'}{v^2} = \frac{x \cdot \frac{1}{x} - \log_{e} x \cdot 1}{x^2} = \frac{1 - log_{e} x}{x^2} = \frac{1 - ln x}{x^2}

\frac{d}{dx}\left(log_{e} x \right) = \frac{1}{x} For stationary points, \frac{dy}{dx} = 0 ie: \frac{1}{x} = 0 multiplying by x gives 1 = 0, which is impossible, \therefore There are no stationary points.

This site knows all and sees all: Australian Tertiary Admission Rank (ATAR) | UAC

your subjects maybe solid, they may be liquid, or gas. Or they even may be plasma or bose-einstein condensates. I suppose there is no way to be sure :S

We have: \sqrt{x} + \sqrt{y} = 2 Note that: x \geq 0\,\, \forall x and y \geq 0\,\, \forall y Futhermore, \sqrt{y} = 2 - \sqrt{x} \left(\sqrt{y}\right)^2 = \left( 2 - \sqrt{x} \right)^2 \therefore y = 4 - 4\sqrt{x} + x The x-intersept occurs when y = 0, hence \sqrt{x} + \sqrt{0} = 2...

Re: Distinction Courses: Interest and discussion on the proposed cancellation of cour I know! I really really want to do Cosmology!

or what about A - TAR?

Re: NSW to dump UAI! WTF! Why would they change it to 99.95? Its like saying "Um... lets give the UAI some weird name and lets just go over to the random number generator and see what number comes out... oh.. its printing now!... the new UAI... i mean STAMT... i mean ATAR... is 504.50! That...

ah, thats seam like a much better way :)

you have to write each line seperately, example (i will exclude the two brakets ie "[tex]" will be "tex") tex x^2 + 2x + 1 /tex tex = (x+1)^2 /tex So essentually you have to write new tex /tex for each line

Well, we are trying to get the function in the form f'(x) ef(x). Notice that the derivative of the exponent ie x^2 is 2x, by we have only x out the front, so if we write it like 1/2 x 2x, which is the same as x, then we can simply do the integration, then half it.

which formulae exactly?

hint: the normal is perpendicular to the tangent, ie m1m2 = -1 hint 2: y - y1 = m(x - x1)

This one is pretty hard (if you dont understand how to achieve the result), so i will just do it for you and you should understand the process. We will proceed by differentiating by first principals: f(x) = 2^x f(x+h) = 2^{x+h} = 2^x \cdot 2^h \frac{dy}{dx} = \lim_{h \rightarrow 0}...

hint: If y = loge f(x), then y' = f'(x) X 1/f(x) = f'(x)/f(x) hint 2: ln is loge !

1. Find the derivative of y = e1 + lnx (hint: y' = f'(x)ef(x)) 2. Remember that the first derivative, ie y' is the gradient of the tangent. 3. Substitute x = e into y' Done!

What kinda series do you think it is, does T_{2}-T_{1} = T_{3} - T_{2}\,\, ? or does \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}}\,\,\,?

Re: maths question all you have to do is substitute x = x+1 into the function f(x+1-1) = (x+1)2 - 1 .: f(x) = x^2 + 2x

$Remembering that ln is the same as$\,\,log_{e} 5 e) \frac{d}{dx}\left(log_{e}\left(7x -2\right) \right) $If$\,\,log_{e}f(x)\,\,$then$\,\, \frac{dy}{dx} = f'(x) \cdot \frac{1}{f(x)} f(x) = 7x - 2 f'(x) = 7 \therefore \frac{d}{dx}\left(log_{e}\left(7x -2\right) \right) = 7 \cdot...

1) The rate of change is the first derivative, so just find that by differentiating p. 2) initial means t = 0, that is, when the time is 0, ie the start. 3)for maximum, the first derivative is zero and the second derivative is negative. Or, as it is a quadratic, you could just find its axis of...