Is this Application of Leibniz Rule in Symbolic Math Toolbox Incorrect?

Referring to Leibniz Integral Rule:

Verify R2025b

matlabRelease

ans = 
  matlabRelease with properties:

    Release: "R2025b"
      Stage: "release"
     Update: 2
       Date: 16-Oct-2025

Define some functions

syms t x
syms f_1(x) f_2(x) g(t)

Case 1 looks correct:

syms G(x)
eqn1 = G(x) == int(g(t),t,f_1(x),f_2(x));
eqn2 = diff(eqn1,x);
case1 = [eqn1;eqn2]

case1 = 

Case 2 looks correct. Same as case 1 with dg(t)/dt replacing g(t).

Dg(t) = diff(g(t),t);
eqn3 = G(x) == int(Dg(t),t,f_1(x),f_2(x));
eqn4 = diff(eqn3,x);
case2 = [eqn3;eqn4]

case2 = 

Case 3 looks correct. Same as case 1 with g(t) = h(t)p(t)

syms h(t) p(t)
eqn5 = G(x) == int(h(t)*p(t),t,f_1(x),f_2(x));
eqn6 = diff(eqn5,x);
case3 = [eqn5;eqn6]

case3 = 

Case 4 looks incorrect

Dp(t) = diff(p(t),t);
eqn7 = G(x) == int(h(t)*Dp(t),t,f_1(x),f_2(x));
eqn8 = diff(eqn7,x);
case4 = [eqn7;eqn8]

case4 = 

I don't understand the meaning of the third multiplicands in both terms of the derivative expression. What does it mean to differentiate with respect to a function?

Seems like case 4 should be:

eqn9 = diff(G(x),x) == h(f_2(x))*Dp(f_2(x))*diff(f_2(x),x) - h(f_1(x))*Dp(f_1(x))*diff(f_1(x),x)

eqn9 = 

Is case 4 (eqn8) incorrect or am I misunderstanding the mathematical meaning of

diff(p(f_2(x)),f_2(x))

ans = 

[Torsten]

I think it's a bad notation that results if you substitute f_1(x) resp. f_2(x) everywhere for t in the functional expressions under the integral.

Mathematically, it doesn't make sense.

But if you take your notation Dp(f_1(x)) resp. Dp(f_2(x)) literally and insert f_2(x) for t everywhere (also for the t in the differential operator d/dt), you get the output from the symbolic toolbox.

[Paul]

I think you're right, but I can't recreate that result using any version of subs, so I don't know how the software gets there.

I'll probably report it as a bug and see what they say.

syms p(t) f_1(x)
subs(diff(p(t),t),t,f_1(x))

ans = 

Dp(t) = diff(p(t),t);
subs(Dp,t,f_1)

ans(t) = 

subs(Dp(t),t,f_1)

ans = 

subs(Dp(t),t,f_1(x))

ans = 

[Torsten]

This doesn't work, either:

syms t x u
syms f_1(x) f_2(x)
syms h(t) p(u)
syms G(x)
Dp(u) = diff(p(u),u);
eqn7 = G(x) == int(h(t)*subs(Dp(u),u,t),t,f_1(x),f_2(x))

eqn7 = 

eqn8 = diff(eqn7,x);
case4 = [eqn7;eqn8]

case4 = 

[Paul]

A work-around for Case 4 is to set it up like Case 3 with a function the represents the derivative of p(t) and then substitute the actual derivative after differentiating the integral.

syms t x
syms f_1(x) f_2(x) G(x) h(t) p(t)
Dp(t) = diff(p(t),t);
syms pdot(t)
eqn7 = G(x) == int(h(t)*pdot(t),t,f_1(x),f_2(x));
eqn8 = diff(eqn7,x);
case4 = [eqn7;eqn8]

case4 = 

eqn9 = subs(eqn8,[pdot(f_1),pdot(f_2)],[Dp(f_1),Dp(f_2)])

eqn9 = 

It looks like the questionable substitution results from something like this

syms z(t)
subs(z(f_1(x)),z,diff(p))

ans = 

[David Goodmanson]

Hi Paul, I think case 4 is correct although it's not the whole story. Using stuff, most of which you're using, like

d/dx (g(f(x)) = g'(f(x)) df(x)/dx
g'(f(x))  = dg(w)/dw | w = f(x)
dg/df = (dg/dx) / (df/dx)

and pretty much ignoring that functions can have singularities and such, you can arrive at what they have. Functional derivative examples:

d(x^3) / d(x^2) = (3x^2 dx) / (2x dx) = (3/2) x
d(cos(theta)) / d(sin(theta)) = -tan(theta)

and so forth.

[Paul]

Hi David,

When I see "functional derivative" I think of Functional Derivative (or in Matlab functionalDerivative), but I don't think that's what you meant.

I understand where you're coming from, but let me make sure.

My claim is that Case 4 result should include the follwing term:

syms p(t) f_1(x)
Dp(t) = diff(p(t),t);
term1 = Dp(f_1)

term1 = 

By the Chain Rule we know that

eqn1 = term1 == diff(p(f_1(x)),x)/diff(f_1(x),x)

eqn1 = 

Interestingly enough, Matlab knows (in accordance with the display) that the RHS is the derivative of p wrt to its argument evaluated at f_1(x) (which is exactly the LHS of eqn1). Note we also saw that notation in Case 2 in my original example. But we don't see that same term displayed in the ouput of Case 4, which might be an indication that something is amiss.

If I'm following correctly, my eqn1 is basically a restatement of your equation:

% dg/df = (dg/dx) / (df/dx)

However, the term that actually shows up in Matlab's result for Case 4 (at least according to its display is)

term2 = diff(p(f_1),f_1(x))

term2 = 

which is not by appearance the same as the LHS or RHS of eqn1 (though maybe it has the same meaning, more on this below).

char(lhs(eqn1))

ans = 'subs(diff(p(t), t), t, f_1(x))'

char(rhs(eqn1))

ans = 'D(p)(f_1(x))'

char(term2)

ans = 'diff(p(f_1(x)), f_1(x))'

Let's put in some actual functions for p(t) and f_1(x)

p(t) = t^2; f_1(x) = x^3;

f_1(x) = 

subs(eqn1) % 2*t evaluated at t = x^3

ans = 

subs(term2)

ans = 

The software itself is apparently unsure of what term2 is, again indicating something might be amiss (but see below).

FWIW, if we go back to the original expression for term2, but try to define it with actual definitions for p(t) and f1(x), we actually get an error

So even if term2 is supposed to be the same as term1, it doesn't seem particularly useful if it can't evaluate when actual expressions for p(t) and f_1(x) are defined (but, more on this below).

In the interest of full disclosure, somehow term2 comes up with the correct result for this f_1(x)

f_1(x) = cos(x);
subs([eqn1;term2])

ans = 

OTOH, we also have this result, which is a bit ... odd

f_1(x) = exp(x);
subs([eqn1;term2])

ans = 

Digging in a little deeper ...

term2 can be defined this way, as long as f_1(x) is not defined

syms p(t) f_1(x)
term2 = diff(p(f_1),f_1(x))

term2 = 

Now define p(t)

p(t) = t^2;
p(f_1)

ans = 

And we get the expected result from diff

diff(p(f_1),f_1(x))

ans = 

or subs

term2 = subs(term2)

term2 = 

Now we can define f_1(x) and get the correct result

f_1(x) = exp(x);
term2 = subs(term2)

term2 = 

So we can get it to work with a two step process.

At the end of the day, I suppose one can argue that my preferred expression and Matlab's actual expression for Case 4 are equivalent

syms p(t) f_1(x)
eqn = subs(Dp(t),t,f_1(x)) == diff(p(f_1),f_1(x))

eqn = 

as long as one understands precisely how to interpret the RHS of eqn and how to work with it after p(t) and f_1(x) are subsequently defined.

Though Matlab doesn't necessarily agree (I show this with a bit of tongue in cheek, because isAlways can be finicky and has to consider corner cases).

isAlways(eqn)

Warning: Unable to prove 'subs(diff(p(t), t), t, f_1(x)) == diff(p(f_1(x)), f_1(x))'.

ans = logical
   0

BTW, this whole exploration started with an expression like this

syms G(x) X h(tau)
eqn = G(x) == int(h(x-tau)*Dp(tau),tau,x-X,x)

eqn = 

diff(eqn,x)

ans = 

When I saw that middle term, I could only guess what it meant from context. IMO, it's not standard mathematical notation.

It's all a bit confusing that Matlab uses three different representations that all (apparently) mean the same thing: lhs(eqn1), rhs(eqn1), and term2.

[David Goodmanson]

Hi Paul,

As you suggested, by 'functional derivative' I didn't mean what is usually meant by that term, i .e. the derivative of a functional (for example an integral with the function as part of the integrand) w/r/t a function. I should have done better with terminology.

Yes, I think the both sides of your eqn 1 and term 2 all represent the same thing, and here is some scribbling on the topic. The answer to the original expression has a lower limit term from f1 and upper limit term from f2. Both are the same problem, so I'll just take the upper limit, call it f(x) and put the lower limit as 0 to get it out of the picture.

Case 4 is basically

G = Int{0, f(x)} h(t) (dp(t)/dt) dt
  = Int{0, f(x)} h(t) p'(t) dt
%  Then
dG/df = h(f) dp(f)/df
      = h(f) p'(f)       

Since dx is not involved, x is merely implicit. f can be considered to be an independent variable, and the fact that it might be a function of x is neither here nor there [except that x might be needed for bookkeeping purposes to specify the value of f = f(x) at the point of interest.]. So it's also true that in that sense,

  dG/df(x) = h(f(x))  dp(f(x)) / df(x)
           = h(f(x))  p'(f(x))       

For the evaluation of the lhs of eqn. 1, for any derivative,

dh(u)/du = dh(y)/dy | y=u

where y is just a dummy variable (t in eqn.1). So all these various expressions are equivalent.

Only with d/dx does x really come into the picture to supply the extra factor from the chain rule.

dG/dx = (dG/df) (df/dx) = h(f) p'(f) df/dx

I should get more serious about syms. In this case it brought up possibilities. Sometimes is works pretty well. In some other cases though, it seems harder to get it to agree with what you know to be true than it is to forego it entirely and work out the result some other way. And it's well known for sometimes cranking out a result that is hundreds of characters long if not thousands, and which might actually be correct.