KSIĄŻKI POWIĄZANE ZE SŁOWEM «LEIBNITZ'S RULE»
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1
Calculus of Finite Differences and Numerical Analysis
Leibnitz's Rule. We have studied Leibnitz's rule for finding n derivative of the
product of two functions u and v as Dn (wv) = nDn(v)+ nCx{Du)Dn~ '(v)+ nC2(D2u
)Dn~ 2(v)+ ...+ nCn(Dnu)v. An analogous formula exists for differences and is
given ...
Prof. P. P. Gupta, Sanjay Gupta, G. S. Malik
2
Higher Engineering Mathematics
Note: Leibnitz's rule is applicable even when one of the limits of integration is
infinite. Worked Out Examples Example 1: Apply Leibnitz's rule £;f_^i eax dx.
Solution: Here /(< rb(a) eax dx ot)= I J-2 = / f(x, a)dx, so b(a) = —a, a(a) Ja(a) —2a
-f(x, ...
3
Linear Difference Equations
We have studied Leibnitz's rule for finding n derivative of the product of two
functions u and v as If (mv) = uD" (v) + "Cj (Dm) D"~1 (v) + nC2 (Z)2w) Z)71~2 (v) .
+ ...+nCn(Dnu)v. An analogous formula exists for differences and is given by An (
uv) ...
Dr. R.K. Gupta, D.C. Agarwal
4
Engineering Mathematics-I (For Wbut)
X y = jf(t)sm[k(x-t)]dt. The upper limit in this integral involves the parameter x. So,
using Leibnitz's Rule, we have X Jx = j~[f(t)sin[k(x-t)}dt o +/(*) sin[*(jc-*)] — (*) dx _
/(0)sin[*(x-0)]^(0) X = / #"(f)cos[*(*-0]*< 0 Using once more the Leibnitz's Rule, ...
5
Engineering Mathematics
sin a. 2 Integrating with respect to a, we get F(a) = — ^cos a + c. But F(0) = 0,
therefore, EXAMPLE 5.86 By successive use of Leibnitz's Rule to fxmdx, J, o
evaluate J ^(logjc)"^. 7T 7T 0 = -- + corc = -. 2 2 Solution. We have Hence, F(<x) =
--cosa ...
Solution: Applying Leibnitz's rule y„(e")n -Inx +nС1(еx)„-\(\nx)\ •.. + nс„(еx)(\пx)„. [
MM. 2005] Solution: Differentiating y w.r.t. x, yi = -a • sin(ln л) • ( - I + b • cos(ln x)- -
\xj x xy\ = —a sin(lnл) + ...
7
Geological Storage of CO2: Modeling Approaches for ...
Leibnitz's rule applies to the derivative of integrals. In the simplest case, we
consider the function h(x,y), which is integrated with respect to its first variable
over a domain bounded by two functions f1(x) and f2(x). Leibnitz's rule now
formalizes ...
Jan Martin Nordbotten, Michael A. Celia,
2011
8
Engineering Mathematics I: For Uptu
So, using Leibnitz's Rule, we have X o +/(*) sm[k{x - x)] — (x) dx _/(0)sin[^-0)]£(0)
X = J kf{t) cos[*(jc - t)]dt. 0 Using once more the Leibnitz's Rule, we get X g = /![*/(,)
cosW*-*, 0 + kf(x) cos[k(x — x)\ — (x) dx -^(0)cos[*(x-0)]£(0). x = -k2 J /(f) sm[k(x ...
Applying Leibnitz's rule with u = sin x, v = x2 d" d"~[ yn = sin , • x + nc , r (sin x) • 2x
dx" ' dx"~' yi = —a • sin(ln x) • I - I + b • cos(ln*) • - \ X f xyi = —a sin(ln x) + b cos(ln
x). Differentiating Xy2 + yi=—a • cos(ln x) • b sin(ln x) • - x x jc2>'2 + xyi ...
10
Demand Uncertainty in International Trade
since exp(-u)g(-u)d-u = 0. dv^.H^ requires use of Leibnitz's Rule: dv2{x*H) _ dx*
H S ^H(x*H)g(x*H) + / expx H Jx* exp(u)g{u)du Since tth(x*h) = 0 and exp(u)g{u)
du > 0 dv2(x*H) = (1-6) f S exp x*H dx H exp(u)g{u)du < 0 dxt, dx* To find —^r^-, ...