Differentiation help please

[gizzemoto]

Hi all, I am a second year student (old timer) and my disciplin is in Computer and electronic enginnering.I am really finding it hard to understand diffeentiation. I can understand the concept but putting this beyond first principles is beyond me. I know i really want to learn this as i know it will come up time after time. If any of you out there can help, i would really appreciate it. Thanks

 

[ericgibbs]

Hi all, I am a second year student (old timer) and my disciplin is in Computer and electronic enginnering.I am really finding it hard to understand diffeentiation. I can understand the concept but putting this beyond first principles is beyond me. I know i really want to learn this as i know it will come up time after time. If any of you out there can help, i would really appreciate it. Thanks

hi,

Look at this practical example.

 

[MrAl]

Hi gizzemoto,

When you first start to get into this it sometimes helps to look at numerical
differentiation first before getting into full blown differentiation.

The numerical derivative of a function is:

f'(x)=(f(x+h)-f(x))/h

where

f(x) is the function
h is a small increment, like 0.001 .

As h gets smaller and smaller, the above equation approaches the
exact derivative at a point x.

As you get going with this, it also helps to look up some common
derivatives like the derivative of x, of x^2, of x^3, etc., and the
formulas that are used to find the derivatives.
For example, the derivative of x^2 is 2*x, and the derivative of
x^3 is 3*x^2, and the general formula for the derivative of
x^N is N*x^(N-1). This isnt too hard to do, but sometimes
the technique is a little more involved so you end up looking
up various methods and sometimes even consult a table
of derivatives. Some math handbooks have extensive tables.

If you would like more help with this, try asking some more
specific questions, like what is the derivative of sin(x) or
something like that. You'll find that eventually you can
find the derivative for almost any function.

 

[Pommie]

When I first tried to learn calculus I found it very confusing. I just didn't understand what I was trying to find. When I discovered that I was trying to find the slope of the graph then dy/dx suddenly made sense.

My wife (mid 40's) is currently doing a refresher course at uni as she is a teacher and I'm explaining lots of this stuff to her at the moment. If you have any examples of stuff you don't understand then post them, I'm sure there are lot's of posters willing to help.

Mike.
Edit, I thought I had better add, in my wifes defense, she hasn't taught for 20 years as she was a Mother to our children. To her credit, she is doing a refresher course rather than go ill equipped into a classroom.

 

[gizzemoto]

Thank you all very much for all your help,

Here is one

7√x In4x I know you use the product rule , but what is the procedure

(2t-1)/(〖3t〗^2+ 5t)

 

[Pommie]

7√x In4x I know you use the product rule , but what is the procedure

If you split it into (7√x) and (ln4X).

Differentiate 7X^½ to get 3.5X^-½

Differentiate ln4X to get 4/4X

Apply the product rule

7√x 4/4X + ln4X 3.5X^-½

Cancel and simplify,

(7+3.5ln4X)/√x

Does that help?

Mike.

 

[HATHA]

hi gizzemoto

d {(2t-1)/((3t))^2 +(5t)}/dt
is this your problem
so i think you know the partial fraction expansion
then then we can write
(2t-1)/((3t))^2 +(5t) =(2t-1)/(9*t^2 +5t)=(2t-1)/t(9t+5)=a/t+b/(9t+5)
where "a" and "b" are real constants
then
(2t-1)/t(9t+5) =(a(9t+5)+bt)/t(9t+5) =[(t(9a+b))+5a]/t(9t+5)
so considering constant term in both side of above equation
we can get
a=-1/5
and considering "t" terms we can get
b=11/5
so we can simplify given function to
d {(2t-1)/((3t))^2 +(5t)}/dt=d {(-1/5t+(11/5(9t+5)) }/dt

d {(-1/5t+(11/5(9t+5)) }/dt= d {(-1/5)t^-1+(11/5)(9t+5)^-1) }/dt
so
it is
d {(2t-1)/((3t))^2 +(5t)}/dt=1/[5(t^2)] - 99/[5(9t+5)^2]
go it

 

[MrAl]

Hi again,

Ok, you know the product rule, so is it the division rule you are after?

For example,

y=3*x^3/(4*x+5*x^2)

and we want to find dy/dx.

The division rule can be remembered like this:

y'=(Ho*dHi-Hi*dHo)/(Ho*Ho)

(pronounced: Ho dee Hi minus Hi dee Ho, over Ho Ho)

where

Hi is the complete numerator,
Ho is the complete denominator,
dHi is the derivative of the numerator,
dHo is the derivative of the denominator

for y=3*x^3/(4*x+5*x^2) we have:

Hi=3*x^3
Ho=4*x+5*x^2
dHi=9*x^2
dHo=10*x+4

Now we form the result using the rule above and get:

dy/dx=((4*x+5*x^2)*(9*x^2)-(3*x^3)*(10*x+4))/(4*x+5*x^2)^2

and after simplification:

dy/dx=(15*x^2+24*x)/(25*x^2+40*x+16)

That's the derivative rule for division.

Of course we could have simplified the expression before
we started to make the problem a little easier.

 

[gizzemoto]

Hi again, Thanks for all your help, Starting to get the hang of it

 

[Willbe]

Speed = d(distance)/d(time). . .?

 

[Salgat]

Differentiation finds the instantaneous slope of a function. Have you tried using tutoring services provided by your school?

 

[gizzemoto]

I would if this was offered, but unfortunately they dont which is sad really, because i know myself and others would benefit greatly

 

[Pommie]

I would if this was offered, but unfortunately they dont which is sad really, because i know myself and others would benefit greatly

Well, hopefully you got some help from the above posts. If you did then maybe if you posted what you gained (now understand) from the above posts then others may benefit.

Do you now have a better understanding of,
the chain rule,
the product rule,
the quotient rule.

If you can explain your thought process then people that read this in the future may get a better understanding of the whole process. If, on the other hand, your still struggling with a particular part of differentiation then just ask.

Mike.

 

[MrAl]

Speed = d(distance)/d(time). . .?

Hi Willbe,


Well, common usage is to call dD/dt 'velocity', which is signed.
Speed, on the other hand, is often taken to exclude the sign,
which removes the directional sense, so that common usage
would mean:

Velocity=dD/dt

and

Speed=abs(Velocity)=abs(dD/dt)

where

abs(x)=|x|=absolute value of x

Also for reference, s is often used for distance instead of D as
used above.

 

[3iMaJ]

It would be sufficient to say that speed is a scalar quantity and velocity is a vector quantity.

 

[MrAl]

It would be sufficient to say that speed is a scalar quantity and velocity is a vector quantity.

Sure, if everyone reading this thread knew what a scalar and a vector was
If not, then i guess it wouldnt be sufficient.

 

[gizzemoto]

An excellent site i came accross is mathtutor.ac.uk and this dose graphical and video lessons on differentation and other topics

 

[Willbe]

Hi Willbe,


Well, common usage is to call dD/dt 'velocity', which is signed.
Speed, on the other hand, is often taken to exclude the sign,
which removes the directional sense, so that common usage
would mean:

Velocity=dD/dt

and

Speed=abs(Velocity)=abs(dD/dt)

where

abs(x)=|x|=absolute value of x

Also for reference, s is often used for distance instead of D as
used above.

I'm going to need some WD-40 'cause I'm rusty.

 

[MrAl]

I'm going to need some WD-40 'cause I'm rusty.

Oh ok, well here's a beer for ya then ---> \~/

 

[Willbe]

Oh ok, well here's a beer for ya then ---> \~/

Your symbol-creating powers are being wasted on this forum. BTW, Pixar Animation is on line 1; should I put them on hold?