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I have
k=(sqrt(2*mw*E0))/hx
q=(sqrt(2*mb*(vp-E0)))/hx
I want to find its derivative w.r.t E0?
I want above derivatives
because I want to find derivative of
(k*tan(k*b/2))-q;
w.r.t E0.
Any one please can help me?
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Hi sonawaneulhas;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym
Thnx a lot for yor help.
Now again i am confused in case of q.
If I am wrong then can you please correct me?
for dq/dE0= -(mb/sqrt(2E0) hx)
Am i right?
And also I dnt knw how to implement these derivatives in
(k*tan(k*b/2))-q;
if you can then can you please help me?
Thanks in advance
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Hi;
And also I dnt knw how to implement these derivatives in
(k*tan(k*b/2))-q;
I don't know what you are trying to do here.
Last edited by bobbym (2009-11-17 21:32:27)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thnx alot bobbym.
Actually my main function is (k*tan(k*b/2))-q;
I am putting values of k and q as first two equations.
After that i have to find out derivative of (k*tan(k*b/2))-q; w.r.t. EO
because in this case my E0 is varying.
got or I am confusing you?
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Hi sonawaneulhas;
All you have to do now is substitute into (k*tan(k*b/2))-q,
with:
k=(sqrt(2*mw*E0))/hx
q=(sqrt(2*mb*(vp-E0)))/hx
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanx alot bobbym.
You have solve my great problem....
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Hi sonawaneulhas;
Glad to help. Let me see what you get when you are all done.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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yes definately...
I will tell u...
all these things i am going to introduce in a programme
Last edited by sonawaneulhas (2009-11-17 22:17:39)
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