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Now u and v cannot be identically zero since if they were, there would be no functional relation, u vx @ u; v contrary to hypothesis. Hence it follows from (1) and (2) that x uy vy @ x; y 0 identically. Su ciency. We have to prove that if the Jacobian @ u; v 0 identically, then there exists a functional @ x; y

smbclient servicename [password] [-s smb.conf] [-B IP addr] \ [-O socket options][-R name resolve order] [-M NetBIOS name] \ [-i scope] [-N] [-n NetBIOS name] [-d debuglevel] [-P] [-p port] \ [-l log file base] [-h] [-I dest IP] [-E] [-U username] \ [-L NetBIOS name] [-t terminal code] [-m max protocol] \ [-W workgroup] [-T<c|x>IXFqgbNan] [-D directory] \ [-c command string] smbclient is a client that can "talk" to an SMB/CIFS server. It offers an interface similar to that of the FTP program. Operations include things like getting files from the server to the local machine, putting files from the local machine to the server, retrieving directory information from the server, and so on.

Tan'''[x] 2 Sec[x]4 + 4 Sec[x]2 Tan[x]2

8.10 Compute the values of the first ten derivatives of f ( x ) = e x at x = 0. Put the results in tabular form.

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relation between u and v, i.e., u; v 0. Let us rst suppose that both ux 0 and uy 0. In this case the Jacobian is identically zero and u is a constant c1 , so that the trival functional relation u c1 is obtained. Let us now assume that we do not have both ux 0 and uy 0; for de niteness, assume ux 6 0. We may then, according to Theorem 1, Page 124, solve for x in the equation u f x; y to obtain x F u; y , from which it follows that 1 u f fF u; y ; yg 2 v gfF u; y ; yg

f[x_] = Exp[x2] derivtable = Table[{n, D[f[x], {x, n}] /. x 0}, {n, 1, 10}]; TableForm[derivtable, TableAlignments Right, TableSpacing {1,5}, TableHeadings {None,{"n", "f(n) (0)"}}]

n 1 2 3 4 5 6 7 8 9 10 f(n) (0) 0 2 0 12 0 120 0 1680 0 30 240

smbd [-D] [-a] [-o] [-d debug level] [-l log file] [-p port number] \ [-O socket options] [-s configuration file] [-i scope] [-P] [h] smbd is the server daemon that provides file sharing and printing services to Windows clients. The server provides file space and printer services to clients using the SMB (or CIFS) protocol\&. This is compatible with the LAN Manager protocol, and can service LAN Manager clients. These include MSCLIENT 3.0 for DOS, Windows for Workgroups, Windows 95, Windows NT, OS/2, DAVE for Macintosh, and smbfs for Linux.

8.11 Sketch the graph of f(x) = x4 50 x2 + 300 and its derivative, on one set of axes, for 10 x 10.

From these we have respectively, 3 4 du ux dx uy dy ux Fu du Fy dy uy dy ux Fu du ux Fy uy dy dv vx dx vy dy vx Fu du Fy dy vy dy vx Fu du vx Fy vy dy From (3), ux Fu 1 and ux Fy uy 0 or (5) Fy uy =ux .

PlotLegends` f[x_] = x4 50 x2 + 300; Plot[{f[x], f'[x]}, {x, 10, 10}, PlotRange { 1000, 1000}, PlotStyle {GrayLevel[0], Dashing[{.015}]}, PlotLegend {"f(x)", "f'(x)"}]

5 500

f(x) f'(x)

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Using this, (4) becomes ux vy uy vx dv vx Fu du fvx uy =ux vy g dy vx Fu du 6 dy: ux @ u; v ux uy But by hypothesis vx vy ux vy uy vx 0 identically, so that (6) becomes dv vx Fu du. @ x; y This means essentially that referring to (2), @v=@y 0 which means that v is not dependent on y but depends only on u, i.e., v is a function of u, which is the same as saying that the functional relation u; v 0 exists.

1000

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