The Geniuses' Thread

Insect Overlord wrote: »

Let c=3 and re-write the equation to suit that?

jumpguy wrote: »

How odd, I was just coming here for maths help myself.

I've a note scribbled in my maths notebook, in a little faraway corner, that the exponential (e) raised to the power of a natural log, say of x (ln x) is equal to x.

i.e e^lnx = x

or, to use a constant

e^ln5 = 5

I've obviously forgotten this, as this was news to me (it's not in any of the T&T books, as far as I know). My calculator is unwilling to verify it too, helpfully letting me know "SYNTAX ERROR". So what I'm wondering is, is this true or is this some mistaken note I scribbled down eons ago?

KnifeWRENCH wrote: »

Use the quotient rule:

[latex]
\displaystyle
\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}
[/latex]

So put u = x -5 and v = x.
Then du/dx and dv/dx both equal to 1.
Now apply the quotient rule and you should end up with 5x^-2. (Unless I missed something, the 1 in the answer you quoted is a typo and shouldn't be there.)

Remember: for any number a, [latex]\frac{1}{x^a} = x^{-a}[/latex]

Btw you could also use the product rule to solve this, by rewriting it as [latex]x^{-1}(x - 5)[/latex]
Or you could just rewrite [latex]\frac{x-5}{x}[/latex] as [latex]1 - \frac{5}{x}[/latex] and differentiate it then.

cocoa wrote: »

suspect knifey's interpeted the question wrong, this is why brackets are the best!

if it's x-(5/x) then the answer given is correct and there's no need for the quotient rule, just the standard rule for differentiating powers. -1 -1 = -2, gives us -1*x^-2, but it was -1/x so we get +5*x^-2

the one is the derivative of x.