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[Skyrunner]

Remember that you can do the following?

log(ab) = log(a) + log(b)

Thus,

ln(5x) = ln(5) + ln(x)

Since ln(5) is a constant, you basically parallelly shifted the graph of ln(x) in the positive y direction, which means the derivative and slope are identical between the two graphs for all x.

[ZetaX]

Log(n) is the order of magnitude of n, whatever base you choose. Thus the statement is equivalent to: the order of magnitude of n and 5n grows by the same if n increases.

[Lightningfalcon]

What is ln^2.  I had it on a quiz Thursday and it's not in the textbook anywheres

[MagmaMcFry]

What is ln^2.  I had it on a quiz Thursday and it's not in the textbook anywheres

I've never seen that before on a ln, but generally f²(x) is kinda ambiguously used as either f(f(x)) or f(x)², and if you say you never saw that notation before, then it's very probably f(x)², since this notation is very common with trigonometric functions ("sin² x" instead of "sin(x)²") and you won't ever find a f(f(x)) in school unless you're practicing derivatives.

[Pnx]

So I'm supposed to use this formula (dy/dx) = 1/(dx/dy) to find the derivative of the inverse of a given function. But I'm kind of confused about how exactly I do this.

[MagmaMcFry]

That's because the formula is written down stupidly. Here's how you derive that formula: Since f(f^-1(x)) = x, the chain rule says that  1 = f(f^-1(x))' = f'(f^-1(x))*(f^-1)'(x), so you get (f^-1)'(x) = 1/f'(f^-1(x)), which is exactly the same thing as the formula you were given, except in actually usable.

For example let f(x) = sqrt(x) for positive x. Then f'(x) = 1/2sqrt(x), f^-1(x) = x², so (f^-1)'(x) = 1/f'(f^-1(x)) = 1/(1/2sqrt(x²)) = 2x.

[Pnx]

So I'm looking at stuff for L'Hopital's Rule and such, and I'm looking at the problem Lim_x->0+ of (cotx)/(lnx). Deriving this doesn't seem like it's going to help at all, and while I remember being told there were some things I could do to get the limit out of this, I don't remember what those things are, and the book doesn't seem to be helpful about this.

Any help?

[MagmaMcFry]

Here's what you do: First you apply L'Hôpital, then you get -csc²(x)/(1/x). Now here's the trick: You switch around the numerator and denominator, rewriting your expression to -x/sin²(x). L'Hôpital that again, and you get -1/2sin(x)cos(x), which visibly converges to negative infinity.

[Pnx]

Ah, thank you, I think I'm starting to get the hang of this.

[Skyrunner]

I am also confused by the inverse derivative thing

[Pnx]

I am also confused by the inverse derivative thing

Hmmm, I don't have the best command of it myself right now, but you might want to be more specific about what's confusing you.

[Skyrunner]

Let's say y1=f(x) and y2=g(x).
The inverse derivative formula is

dx/dy = 1/ (dy/dx)

It looks like the derivative of g(x) is the derivative of f(x), except you swap x and y and represent it in terms of x. But that's not only imoossible for me to do for some equations (x=y^3+3y), it's also incorrect when I do do it.

[Bouchart]

Let's say you have a X by X grid of squares.  In each square, you can place an A or a B.  Each has the following value:

A is always worth 1 point.
B is worth 1 point plus one for each A in a square adjacent or diagonal to it.

Define the value of an X by X grid of squares as the sum of all A's and B's within it.

So if we look at a simple 3 by 3 grid with the following values:

A B A
B A A
A A B

We can see that it is worth:

6 (for the value of the A's)
plus 5 (for the value of B in the first row)
plus 5 (for the value of B in the second row)
plus 4 (for the value of B in the third row)

for a total of 20.

Is there a general rule to find a maximum value of an X by X grid?

[Virex]

According to the chain rule, for any x, y and z:
dy/dx dx/dz = dy/dz

Additionally, for any x,
dx/dx = 1

Now we can make the following derivation:

1 = dx/dx
dx/dx = dx/dy dy/dx
1 = dx/dy dy/dx

Then, if dy/dx != 0
dx/dy = 1/(dy/dx)

[MagmaMcFry]

Skyrunner:

Let's say that g is the inverse of f. In your example (namely f(x) = x³+3x) this g is not really expressible in a way that is not huge: in this case,
  g(y) = (y/2 + (y^2/4 + 1)^(1/2))^(1/3) - 1/(y/2 + (y^2/4 + 1)^(1/2))^(1/3)
  and g'(y) = (1/6)*2^(1/3)/(y^2 + 4)^(1/2)/(y + (y^2 + 4)^(1/2))^(1/3)*(2^(1/3)*(y + (y^2 + 4)^(1/2))^(2/3) + 2),
and you can't be expected to actually calculate that in an exam, so the question was probably posed differently than "Calculate the inverse derivative of f(x) = x³+3x as a function of y". Am I correct in this assumption?