D[1, 2](f)(x, y) means diff(f(x,y), x, y). It means first derivative in first coordinate, first derivative in second coordinate. Your expression D[1, 2](1/x); implies x is a function of 2 coordinates.

These problems tend to be difficult for obvious reasons. See the following:

E0 := x^2 + y^2 - 10*x - 75 = 0;
E := (a*x + b*y + c)^2 + (d*x + e*y + f)^2 = r^2;
V := [[a, b, c], [d, e, f], r];

(lhs - rhs)(expand(E - E0));
C := {coeffs}(%, {x, y});
T := eliminate(C, {c, f, r});
S1 := simplify(T[1], T[2]);
S2 := [solve](T[2], {a, b, d, e});

#A solution:
applyop(evala, 2, subs(S1, S2[1], E));
subs(S1, S2[1], V);

#Another method:
map2(remove, evalb, evala({solve}(C, {a, b, c, d, e, f, r})));
select(proc(S) subs(S, r); is(0 < %) = true; end proc, %);
map(factor, map(allvalues, %));
subs(b=1, %);

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8*x*exp(-3*x^2) - 24*x^3*exp(-3*x^2);
factor(%);
solve(%, {x});

[sin(x), cos(x), tan(x), sec(x), csc(x), cot(x), sinh(x), cosh(x), tanh(x), sech(x), csch(x), coth(x)];
map(f -> f-convert(f, exp), %);
simplify(%);

Simplify does not reach across the exp & trig boundary. You are expected to introduce the relevant identities yourself.

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Here is a procedure tha draws a cylinder from a specified base, along a vector and of a given radius and number of sides:

C := proc(d::list, v::list, r, n::posint)
local x, y, z, f;
description `Draws a cylinder with base at d along vector v, radius r, having n sides. Attach additional options for the plottools[cylinder] command used.`;

use LinearAlgebra in applyop(Normalize, {2, 3}, GramSchmidt([Vector(v), op(NullSpace(v))]), 2); end;
f := unapply(convert(add((% *~ [z, x, y])) + Vector(d), list), [x, y, z]);
use plottools in transform(f)(cylinder([0, 0, 0], r, 1, strips = n, _rest)); end;
end proc;

plots[display](seq(C([0, 0, 1], [10, 3, 10], 1, 40, op(opts)), opts = 
[[color = COLOUR(RGB, 0.2, 0.4, 0.2), transparency = 0, style = contour, thickness = 0, linestyle = 1], 
[color = green, transparency = 0.8, style = patchnogrid]]), 
scaling = constrained, orientation = [160, 45, 30], contours = 20, labels = ['x, y, z'], lightmodel = light4);

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F:=f->taylor(f, x, 4);
F(ln(1+x));
F(ln(1+sin(x)));
F(%%-%);
 

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Like before:

A := cos(5*t)=a*cos(t)^5+b*cos(t)^3*sin(t)^2+c*cos(t)*sin(t)^4;
collect(combine((lhs-rhs)(A)), {sin, cos});
solve({coeffs}(%, indets(%, dependent(t))));

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How is this different from this question you asked?
https://www.mapleprimes.com/questions/228216-Question-From-Maple

mul(1/(i+j),j=0..3);
sum(%,i=1..infinity)=1/18;

Try:

plots[pointplot]( [seq([i,S[i]],i=1..numelems(S))], style=line);

You are missing a condition, so it acts like a parameter.

Example:

dsys := {diff(s(x), x, x, x)+(1/2)*s(x)*(diff(s(x), x, x)) = 0, s(0) = 0, (D(s))(5) = 1};
other:=seq({D(s)(0)=i/10}, i=-8..1);;
sol:=seq(dsolve(dsys union cond,numeric), cond=[other]);
plots[display](map(plots[odeplot], [sol], 0..5));

restart;
PDEtools[declare](f(x), prime = x);

N := 4;
F := sum(p^i*f[i](x), i = 0 .. N);
HPMEq := (1 - p)*(diff(F, `$`(x, 3))) + p*((diff(F, `$`(x, 3))) + 1/2*(diff(F, x, x))*F);
for i from 0 to N do equ[2][i] := coeff(HPMEq, p, i) = 0 end do;

cond[1][0] := f[0](0) = 0, (D(f[0]))(0) = 0, (D(f[0]))(5) = 1;
for j to N do cond[1][j] := f[j](0) = 0, (D(f[j]))(0) = 0, (D(f[j]))(5) = 0 end do;
for i from 0 to N do dsolve({cond[1][i], equ[2][i]}, f[i](x)); assign(%) end do;

g := evalf@unapply(simplify(sum(f[n](x), n = 0 .. N)),x);
convert(g(x), 'rational');  
subs(x = 2.4, diff(g(x), x));


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convert(A,set):
indets(%,name);
fsolve(%%, % , (x->(x=0..1))~(%));

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Try one of these :

Button("Deduct", Evaluate(f = '[work(3), work(2)]'));
Button("Deduct", Evaluate(f = 'proc() work(3);  work(2) end()'));

How about this:

2.55*10^(-90);
op(%);
%[1]/10^Digits;
evalf(%,3);