If you think  diff  command is  too long to write, in 1d input you can use  D  command instead:

restart;
f:=x->x^2;
D(f);
D(f)(x);

# D command is especially useful if you need to calculate the derivative at a point, for example

D(f)(3);

See the plot:

plot(arccos(cos(phi)), phi=0..4*Pi, scaling=constrained);

To get  0  you need to restrict  phi :

cc:=arctan(sin(phi)/cos(phi))-arccos(cos(phi));      
simplify(cc) assuming phi>0 and phi<Pi/2;

Addition. Here are 2 exercises for students (should be solved manually) and to  check in Maple: to calculate  arccos(cos(20)), arccos(sin(20))  (20 radians rather than degrees)
 

Add the line  simplify(Cr);

It's not a bug. In Maple there are cases when the calculation is not performed immediately, but some additional commands are required. Here are 2 examples:

eval(sqrt(a), a=4);
simplify(%);
cos(2*arccos(1/3));
expand(%);
 

Use  plottools:-getdata  command.

P:=a*s^3+b*s^2+c*s+d;
n:=degree(P, s);
add(C||k*s^k, k=0..n);

The example:

convert(F32, string);
parse(substring(%, 2..-1));
                                                    "F32"
                                                       32

# Or

parse(convert(F3, string)[2..-1]);

You cannot mix a variable and an indexed variable with the same name in the same code. Take a look:

a:=1; a[1]:=2;
a+a[1];
                                                 

In your code, exactly the same situation with  r  and  r[1] .

restart:
with(DETools):
with(PDEtools):
u[o](r,z):=(-1/4)*diff(p[o](z),z)*(1-r^2):
ode:=gamma1*diff(u[o](r,z),z)+(1/r)*diff(v[1](r)*r,r)=0:
IC1 := v[1](0) = 0:
ans2 := dsolve({ode, IC1}, v[1](r));

The built-in command  diff  does not allow you to search for the derivative of a function by function, but if you use the well-known chain rule to differentiate the composition of functions, then we get:

q := [q1(t),q2(t),q3(t)];
L:=cos(q1(t))+sin(q2(t))+5*diff(q1(t),t) + 4*diff(q3(t),t);
diff(diff(L,t)/diff(q1(t),t), t);
                   

Edit.

h1 := solve(Vdc = 0.1500000000e-2*sqrt(2.53669508*10^8*u^3-6.06101011*10^8*u^2+3.46343435*10^8*u), u): 
A := plot([h1], Vdc = 0 .. 11.5, color = [magenta], thickness = 1): 
B := plot(Vector([0, 3.38, 5.21, 6.97, 8.4108, 10.099, 10.9232, 11.8091]), Vector([0, 0.760e-1, .1275, .1994, .2286, .3222, .3637, .999]), style = point, symbol = asterisk, color = "Blue"):
plots:-display(A, B);

For example:

restart;

eval(x+y, {x=13, y=259});

# or

assign({x=13, y=259});
x+y, x*y;
 

Should be   lambda:=3;  not  lamba:=3  in the line after  restart;

Use an inert form to avoid automatic simplification:

a%^(-3/4);
                            

restart;
plot3d(x^2 + y^2, x=0..1, y=0..x, filled=true, grid=[20,20]);

# or

plots:-shadebetween(0, x^2 + y^2, x=0..1, y=0..x, grid=[20,20]);

We see that  plots:-shadebetween   also works.
 

restart;
d1:=3: d2:=3: d3:=5:
plots:-implicitplot3d(max(x^2+y^2+z^2-d1^2, (x-3)^2+y^2+z^2-d2^2, x^2+y^2+(z-4)^2-d3^2), x=0..4, y=-3..3, z=-2..3, style=surface, color=khaki, numpoints=500000, axes=normal, scaling=constrained);
               

The user lighting was used.

Addition. We can build everything together in one plot:

restart;
d1:=3: d2:=3: d3:=5:
A:=plots:-implicitplot3d(max(x^2+y^2+z^2-d1^2,(x-3)^2+y^2+z^2-d2^2,x^2+y^2+(z-4)^2-d3^2), x=0..4,y=-3..3,z=-2..3, style=surface, color=khaki, numpoints=500000, axes=normal, scaling=constrained):    
B:=plots:-implicitplot3d([x^2+y^2+z^2=d1^2,(x-3)^2+y^2+z^2=d2^2,x^2+y^2+(z-4)^2=d3^2],x=-5..6,y=-5..5,z=-4..9,color=["LightPink","LightGreen","LightBlue"], style= wireframe, thickness=0,  numpoints=5000):
plots:-display(A, B);