Random Maths Questions! (2 Viewers)

[Coookies]

So I'm revising....

How do you differentiate y=-3/x to y''?
I need to know the steps!

 

[Timske]

Y = -3/x = -3x^-1
Y' = 3x^-2
Y" = -6x^-3

 

[Coookies]

I know the answers but what did you say to yourself in your head when you did it? Like "move 3 to the front etc" (I know thats not what you do lol)

 

[zeebobDD]

I know the answers but what did you say to yourself in your head when you did it? Like "move 3 to the front etc" (I know thats not what you do lol)

lol i say bring the power down and subtract original power by 1:L

 

[Coookies]

lol i say bring the power down and subtract original power by 1:L

I don't think that quite applies to this equation lol

 

[Peeik]

Question: y=-3/x

Answer:
Step 1: Write the question in index form i.e. y =-3x^-1
Step 2: Bring the power down and multiply the coefficient. Also remember to subtract one from the original power. i.e. 3x^-2
Step 3: To differentiate again, bring down the power down and multiply the coefficient. Also remember to subtract from from the original power. i.e. -6x^-3

 

[zeebobDD]

I don't think that quite applies to this equation lol

then you clearly havent understood the equation because it does.

 

[Timske]

Question: y=-3/x

Answer:
Step 1: Write the question in index form i.e. y =-3x^-1
Step 2: Bring the power down and multiply the coefficient. Also remember to subtract one from the original power. i.e. 3x^-2
Step 3: To differentiate again, bring down the power down and multiply the coefficient. Also remember to subtract from from the original power. i.e. -6x^-3

^ my bad i should have explained my steps

 

[such_such]

Another method:

<a href="http://www.codecogs.com/eqnedit.php?latex=y=\frac{-3}{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=\frac{-3}{x}" title="y=\frac{-3}{x}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=y'=-3 \cdot \frac{1}{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=-3 \cdot \frac{1}{x}" title="y'=-3 \cdot \frac{1}{x}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=y'=-3 \cdot -\frac{1}{x^{^{2}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y'=-3 \cdot -\frac{1}{x^{^{2}}}" title="y'=-3 \cdot -\frac{1}{x^{^{2}}}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=y'=\frac{3}{x^{^{2}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y'=\frac{3}{x^{^{2}}}" title="y'=\frac{3}{x^{^{2}}}" /></a>

then
<a href="http://www.codecogs.com/eqnedit.php?latex=y'={3}{x^{^{-2}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y'={3}{x^{^{-2}}}" title="y'={3}{x^{^{-2}}}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=y''={-6}{x^{^{-1}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y''={-6}{x^{^{-3}}}" title="y''={-6}{x^{^{-1}}}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=y'=-6x^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y'=-6x^3" title="y'=-6x^3" /></a>
But you will have to, in the future, look at an equation and expect to solve it by just looking at it.

 

[AAEldar]

in this form you could use the quotient rule (however that is not advised).

much better form, then do as Peeik said.

 

[TimeGoesTooFast]

A quick method that truong has taught me is : (1/v)'

You prove it via the quotient rule and you should end up with -v' / v^2 <- that should be the formula you remember for these type of questions.

Hope it helps And goodluck ^^

 

[Carrotsticks]

A quick method that truong has taught me is : (1/v)'

You prove it via the quotient rule and you should end up with -v' / v^2 <- that should be the formula you remember for these type of questions.

Naughty student! Bad!

Also, did you mean 'chain rule'? Quotient rule can be used I suppose but for something like this, Chain rule is preferable.

 

[TimeGoesTooFast]

Naughty student! Bad!

Also, did you mean 'chain rule'? Quotient rule can be used I suppose but for something like this, Chain rule is preferable.

Lol what? hehe :W Oh and I meant use quotient rule to prove that formula.

If I was doing it, I'd just drag the power down

 

[Autonomatic]

Lol what? hehe :W Oh and I meant use quotient rule to prove that formula.

If I was doing it, I'd just drag the power down

lOl learning and memorising the differentiation formulas and playing around with them is fun and easy... until it comes to application

 

[SpiralFlex]

Explanation editing.

- First consider a few things. This situation is analogous to my hate for CityRail. Pretend I am having an argument with a train driver. The train driver in this case is the number -3. Spiral (Spirally exponent is the 1)

- The vinculum is just a horizontal line that mathematically groups an expression. Here this represents the wall. When the wall is released, the differentiation process begins. (Argument begins)

- So I begin the argument with the driver. I hit him via the process of multiplication, so -3 is multiplied by 1. The sign also changes.

- Since I have beaten up the train driver, I get one Spiral exponent in the form of a cookie! So add an exponent to the bottom part of the fraction.

Another example!

 

[Carrotsticks]

What the...

 

[SpiralFlex]

What the...

Sorray, I'm on sugar.

 

[TimeGoesTooFast]

What the...

haha. Nice reaction

And yeah, Cityrail has stuffed up A LOT

 

[SpiralFlex]

Due to CityRail my spot is no longer reserved due to overcrowding. I use to be able to claim one spot back a long time ago...

 

[Timske]

cookie = -78/x^14

ok