Differential

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• Mar 26th 2007, 01:59 PM
qbkr21
Differential
I was just wanting to know if some could show me how to work one of these:

y = 3 x^2 + 3 x + 3

Find the differential dy when x = 3 and dx = 0.4

thus

y' = 6x+3

If someone could just one out I could pick on the pattern and finish the rest of them.

Thanks!
• Mar 26th 2007, 03:08 PM
topsquark
Quote:

Originally Posted by qbkr21
I was just wanting to know if some could show me how to work one of these:

y = 3 x^2 + 3 x + 3

Find the differential dy when x = 3 and dx = 0.4

thus

y' = 6x+3

If someone could just one out I could pick on the pattern and finish the rest of them.

Thanks!

Another format for the derivative is:
y' = dy/dx, so:
dy/dx = 6x + 3

Now "multiply" both sides by dx:
dy = (6x + 3)dx

Now just plug the numbers in.

-Dan

PS The "multiplication" I did really isn't a multiplication. It's a looooong story. Suffice it to say we can get away with it, but have a care to check the truth of it before you do it.
• Mar 26th 2007, 03:36 PM
qbkr21
Re:

Example:

If y = 5sqrt(x)

Find the change in Y, delta y when x = 1 and delta x = 0.3

I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

:eek:
• Mar 26th 2007, 04:11 PM
topsquark
Quote:

Originally Posted by qbkr21

Example:

If y = 5sqrt(x)

Find the change in Y, delta y when x = 1 and delta x = 0.3

I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

:eek:

The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
dy = 5*(1/2)*1/sqrt(x) dx

(Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

-Dan
• Mar 26th 2007, 04:28 PM
qbkr21
Re:
Dan You Are By God The Man!!!!!!!!!!!:) :)
• Mar 26th 2007, 04:35 PM
topsquark
Quote:

Originally Posted by qbkr21
Dan You Are By God The Man!!!!!!!!!!!:) :)

(I'm blushing!) You're welcome!

-Dan
• Mar 26th 2007, 05:50 PM
qbkr21
Quote:

Originally Posted by topsquark
The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
dy = 5*(1/2)*1/sqrt(x) dx

(Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

-Dan

However when I am working this problem it gives me the same answer we got before .750. I think that the answer must be something else...
• Mar 26th 2007, 05:53 PM
Jhevon
Quote:

Originally Posted by qbkr21
I think that the answer must be something else...

• Mar 26th 2007, 05:57 PM
qbkr21
Re:
Webworks the online homework software does:

Here are the two questions maybe this will make it a bit clear. Thank You for your help!

Here are the 2 Problems (they are totally separate):

http://item.slide.com/r/1/99/i/x5zfX...0iro7gIOrNVPG/
• Mar 26th 2007, 06:04 PM
Jhevon
Quote:

Originally Posted by qbkr21
Webworks the online homework software does:

Here are the two questions maybe this will make it a bit clear. Thank You for your help!

Here are the 2 Problems (they are totally separate):

ok, so i don't see why our method didn't work for the second one. i'll think about it. but lets work on the first.

y = ln{sqrt[(6x + 5)/(7x - 4)]} ..........let's simplify this monstrosity a bit

=> y = (1/2)ln[(6x + 5)/(7x - 4)]
=> y = (1/2)ln(6x + 5) - (1/2)ln(7x - 4) ...........now let's differentiate
=> dy/dx = 3/(6x + 5) - (7/2)/(7x - 4)

hold on, what did you want to do with this problem?
• Mar 26th 2007, 06:07 PM
topsquark
Quote:

Originally Posted by topsquark
The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
dy = 5*(1/2)*1/sqrt(x) dx

(Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

-Dan

Quote:

Originally Posted by qbkr21
However when I am working this problem it gives me the same answer we got before .750. I think that the answer must be something else...

x = 1, (Delta)x = 0.3

(Delta)y = 5*(1/2)*1/sqrt(1)*0.3 = 0.75

is what I get. I can't see what else it could be.

-Dan
• Mar 26th 2007, 06:09 PM
qbkr21
Re:
I mean the big LN out in front confuses me I know that:

(1/2)ln

but then to I recopy the problem and log the 2nd half this, and this is the point in which I get confused. If the big LN were not there I would

(1/2)[(6x+5)/(7x-4)^(-1/2) then I would differentiate the inside using the quotient rule for the inside of the square root. I really am stuck Jhevon. :o
• Mar 26th 2007, 06:15 PM
qbkr21
Re:
Here is the full problem this is just what it looks like on the computer:

http://item.slide.com/r/1/126/i/RbKf...XF6i2cz3V8xZ-/
• Mar 26th 2007, 06:15 PM
Jhevon
Quote:

Originally Posted by qbkr21

Example:

If y = 5sqrt(x)

Find the change in Y, delta y when x = 1 and delta x = 0.3

I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

:eek:

see if this solution works:

y = 5sqrt(x)

now, (delta)y = f(x + (delta)x) - f(x)

now f(x + deltax) = f(1 + 0.3) = f(1.3) = 5.700877125

f(x) = f(1) = 5

=> deltay = 5.700877125 - 5 = 0.700877125
• Mar 26th 2007, 06:16 PM
topsquark
Quote:

Originally Posted by qbkr21
I mean the big LN out in front confuses me I know that:

(1/2)ln

but then to I recopy the problem and log the 2nd half this, and this is the point in which I get confused. If the big LN were not there I would

(1/2)[(6x+5)/(7x-4)^(-1/2) then I would differentiate the inside using the quotient rule for the inside of the square root. I really am stuck Jhevon. :o

You know what the ln function is yes? Well d/dx(ln(x)) = 1/x. Then you can use the chain rule from there, just like Jhevon did.

-Dan
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