# Math Help - Differential

1. ## 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!

2. 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.

3. ## 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!

4. 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!

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

5. ## Re:

Dan You Are By God The Man!!!!!!!!!!!

6. Originally Posted by qbkr21
Dan You Are By God The Man!!!!!!!!!!!
(I'm blushing!) You're welcome!

-Dan

7. 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...

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

9. ## 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):

10. 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?

11. 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
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

12. ## 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.

13. ## Re:

Here is the full problem this is just what it looks like on the computer:

14. 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!

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

15. 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.
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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