Simplify outputs huge result for diff symbolic expression with hyperbolic functions

Matlab R2019 Update 3

syms x y k real;
syms f(x,y);
syms a b;

xd = -a*tanh(k*f)+b*sech(k*f);
yd = -b*tanh(k*f)-a*sech(k*f);

chid = atan2(yd, xd);

d_chid_dx = diff(chid, x);

simplify(d_chid_dx, 'Steps', 100, 'Criterion', 'preferReal')

produces result

ans(x, y) =
 
(((real(b*k*diff(f(x, y), x)*(tanh(k*f(x, y))^2 - 1)) - imag((b*k*sinh(k*f(x, y))*diff(f(x, y), x))/cosh(k*f(x, y))^2) + real((a*k*sinh(k*f(x, y))*diff(f(x, y), x))/cosh(k*f(x, y))^2) + imag(a*k*diff(f(x, y), x)*(tanh(k*f(x, y))^2 - 1)))/(imag(b*tanh(k*f(x, y))) - real(a*tanh(k*f(x, y))) + imag(a/cosh(k*f(x, y))) + real(b/cosh(k*f(x, y)))) - ((imag((a*k*sinh(k*f(x, y))*diff(f(x, y), x))/cosh(k*f(x, y))^2) - real(a*k*diff(f(x, y), x)*(tanh(k*f(x, y))^2 - 1)) + real((b*k*sinh(k*f(x, y))*diff(f(x, y), x))/cosh(k*f(x, y))^2) + imag(b*k*diff(f(x, y), x)*(tanh(k*f(x, y))^2 - 1)))*(imag(a*tanh(k*f(x, y))) + real(b*tanh(k*f(x, y))) - imag(b/cosh(k*f(x, y))) + real(a/cosh(k*f(x, y)))))/(imag(b*tanh(k*f(x, y))) - real(a*tanh(k*f(x, y))) + imag(a/cosh(k*f(x, y))) + real(b/cosh(k*f(x, y))))^2)*(imag(b*tanh(k*f(x, y))) - real(a*tanh(k*f(x, y))) + imag(a/cosh(k*f(x, y))) + real(b/cosh(k*f(x, y))))^2)/((imag(a*tanh(k*f(x, y))) + real(b*tanh(k*f(x, y))) - imag(b/cosh(k*f(x, y))) + real(a/cosh(k*f(x, y))))^2 + (imag(b*tanh(k*f(x, y))) - real(a*tanh(k*f(x, y))) + imag(a/cosh(k*f(x, y))) + real(b/cosh(k*f(x, y))))^2)

while calculation by hand gives

-a*k*sech(k*f(x,y))

which is correct.

I tried options for simplification (rewrite, steps, criterion) but with no luck.

Any answer will be greatly appreciated.

Answer 1

[Steven Lord]

Are you sure about your hand-calculated answer? Let's substitute one simple f(x, y) function into both the answer from Symbolic Math Toolbox and your hand-calculated answer and compare the results. I've run the code you wrote above before running any of these commands.

answer2 = -a*k*sech(k*f(x, y));

subs(yd, f, 0) % yd with f(x, y) = 0
subs(xd, f, 0) % ditto for xd

The values of xd and yd with this substitution are constants. Therefore so is chid, it has no dependency on x or y. The derivative of a constant should be 0.

subs(d_chid_dx, f, 0) % It is 0

What happens when we substitute f(x, y) = 0 into your answer?

subs(answer2, f, 0) % -a*k

As other checks, substituing f(x, y) = x into both d_chid_dx and answer2 gives answers that differ by a factor of a.

>> subs(answer2, f(x, y), x)
ans =
-(a*k)/cosh(k*x)

>> simplify(subs(d_chid_dx, f(x, y), x))
ans(x, y) =
-k/cosh(k*x)

Substituting f(x, y) = y also gives different answers.

>> subs(answer2, f(x, y), y)
ans =
-(a*k)/cosh(k*y)

>> simplify(subs(d_chid_dx, f(x, y), y))
ans(x, y) =
0

For this last check, you're taking the derivative with respect to x of a function that doesn't include x at all so it seems reasonable that the result ought to be 0.

[Steven Lord]

Using the information from the follow-up, I told MATLAB that a, b, and c are all real.

syms x y k real;
syms f(x,y);
syms a b c real;

f(x, y) = a*x + b*y + c;
xd = -a*tanh(k*f)+b*sech(k*f);
yd = -b*tanh(k*f)-a*sech(k*f);
chid = atan2(yd, xd);
d_chid_dx = diff(chid, x);

>> simplify(d_chid_dx)
ans(x, y) =
-(a*k)/cosh(k*(c + a*x + b*y))

That is in agreement with your result, albeit expressed in terms of 1/cosh rather than sech.

>> answer2 = -a*k*sech(k*f(x, y));
>> answer1 = simplify(d_chid_dx);
>> isAlways(answer1 == answer2)
ans =
  logical
   1

[Andrey Mironenko]

Great, thanks! So the point was to specify that a and b are also real and not just x and y.

Answer 2

[Andrey Mironenko]

Yes I am sure. Sorry, my bad and thanks for the good catch, but I forgot to mention that in my particular case

 f(x,y) = a*x + b*y + c

so

diff(f, x) = a

and

diff(f,y) = b

and therefore if

f(x,y) = 0

then

a = 0

and

b = 0

and so

d_chid_dx = 0

just as you noticed. I should have pointed out this precisely of course. So the problem remains.