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Display derivative with prime

Setup

To better focus on the actual question, in this answer we

  • use simpler names: w for w0, V for EU
  • skip the assumption that $p \in [0, 1]$

Define U (as a symbolic function U, not as the expression U(x)):

sage: U = function('U')

Define a function V of three variables based on U:

sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)

Note that all names on the left hand side of the assignment operator = get defined by that line: V as a function, and x, y, p as symbolic variables. No need to declare them beforehand.

Define three symbolic variables:

sage: w, g, c = SR.var('w, g, c')

Define an expression f based on V and these variables:

sage: f = V(w, g*w - c, p)

Differentiate with respect to w:

sage: h = diff(f, w)

View the result:

sage: view(h)

$$ -g {\left(p - 1\right)} \mathrm{D}_{0}\left(U\right)\left(g w - c\right) + p frac{partial}{partial w}U\left(w\right) $$

That has two occurrences of $U'$, represented in two ways:

  • $U'(g w - c)$ is represented as $\mathrm{D}_{0}\left(U\right)\left(g w - c\right)$,
  • $U'(w)$ is represented as $\frac{\partial}{\partial w} U(w)$

The goal is to make them appear as $U'(g w - c)$ and $U'(w)$.

There are two approaches:

  • use replacements in the resulting latex string
  • use substitutions in the expression tree

The answer by @dsejas covers string replacements in the latex form.

Here we discuss substitutions in the expression tree.

First we give a ready-made solution. Then we explain how one could explore to get to that solution.

Make a particular derivative display with prime

Here is a function for operator substitution in an expression:

def subs_op(expr, op1, op2):
    r"""
    Return the expression `expr` with operator `op1` replaced by `op2`.
    """
    op = expr.operator()
    if op:
        opp = (op2 if str(op) == str(op1) else op)
        args = expr.operands()
        aargs = (subs_op(a, op1, op2) for a in args)
        return opp(*aargs)
    else:
        return expr

To address the question with this, define the to-be-replaced and the replacement operators, then use subs_op:

sage: DU = U(x).diff().operator()
sage: Uprime = function("U'")
sage: fff = subs_op(ff, DU, Uprime)
sage: fff
-g*(p - 1)*U'(g*w - c) + p*U'(w)
sage: view(fff)

$$ -g {\left(p - 1\right)} {\rm U'}\left(g w - c\right) + p {\rm U'}\left(w\right) $$

Problem solved, at least in an ad-hoc way.

Make all derivatives display with prime

Here is a use_prime function that walks through an expression and replaces any derivatives of functions of a single variable by "prime" versions (or "double-prime", etc. depending on the order of differentiation).

def use_prime(expr):
    r"""
    Return the expression with one-variable function derivatives "primed".
    """
    op = expr.operator()
    if op:
        args = expr.operands()
        aargs = (use_prime(a) for a in args)
        opp = op
        if (isinstance(op, sage.symbolic.operators.FDerivativeOperator)
                and len(args) == 1):
            name = op.function().name()
            primes = "'" * len(op.parameter_set())
            opp = function(f"{name}{primes}")
        return opp(*aargs)
    else:
        return expr

Example with various types of derivatives:

sage: a, b = (function(f) for f in 'ab')
sage: x, z = SR.var('x, z')
sage: f(x, z) = z * a(x * z) + x * z * diff(b(z), z)
sage: g(x, z) = diff(f(x, z), x) + diff(f(x, z), z)
sage: h = diff(g(x, z), x)
sage: h
x*z^2*D[0, 0](a)(x*z) + z^3*D[0, 0](a)(x*z) + 2*z*D[0](a)(x*z) + z*diff(c(z), z, z) + diff(c(z), z)
sage: view(h)

$$ x z^{2} \mathrm{D}_{0, 0}\left(a\right)\left(x z\right) + z^{3} \mathrm{D}_{0, 0}\left(a\right)\left(x z\right) + 2 \, z \mathrm{D}_{0}\left(a\right)\left(x z\right) + z frac{partial^{2}}{(partial z)^{2}}c\left(z\right) + frac{partial}{partial z}c\left(z\right) $$

sage: hh = use_prime(h)
sage: hh
x*z^2*a''(x*z) + z^3*a''(x*z) + 2*z*a'(x*z) + z*c''(z) + c'(z)
sage: view(hh)

$$ x z^{2} {\rm a''}\left(x z\right) + z^{3} {\rm a''}\left(x z\right) + 2 \, z {\rm a'}\left(x z\right) + z {\rm c''}\left(z\right) + {\rm c'}\left(z\right) $$

Example with the original question:

U = function('U')
EU(x, y, p) = p*U(x) + (1 - p)*U(y)
w0, g, c = SR.var('w0, g, c')
f = EU(w0, g*w0 - c, p)
h = diff(f, w0)
show(h)
hh = use_prime(h)
show(hh)

$$ -g {\left(p - 1\right)} \mathrm{D}_{0}\left(U\right)\left(g w_{0} - c\right) + p frac{partial}{partial w_{0}}U\left(w_{0}\right) $$

$$ -g {\left(p - 1\right)} {\rm U'}\left(g w_{0} - c\right) + p {\rm U'}\left(w_{0}\right) $$

Display derivative with prime

Setup

To better focus on the actual question, in this answer we

  • use simpler names: w for w0, V for EU
  • skip the assumption that $p \in [0, 1]$

Define U (as a symbolic function U, not as the expression U(x)):

sage: U = function('U')

Define a function V of three variables based on U:

sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)

Note that all names on the left hand side of the assignment operator = get defined by that line: V as a function, and x, y, p as symbolic variables. No need to declare them beforehand.

Define three symbolic variables:

sage: w, g, c = SR.var('w, g, c')

Define an expression f based on V and these variables:

sage: f = V(w, g*w - c, p)

Differentiate with respect to w:

sage: h = diff(f, w)

View the result:

sage: view(h)

$$ -g $$-g {\left(p - 1\right)} \mathrm{D}_{0}\left(U\right)\left(g 1\right)} {\rm D}_{0}\left(U\right)\left(g w - c\right) c\right) + p frac{partial}{partial w}U\left(w\right) $$w}U\left(w\right)$$

That has two occurrences of $U'$, represented in two ways:

  • $U'(g w - c)$ is represented as $\mathrm{D}_{0}\left(U\right)\left(g w - c\right)$,
  • $U'(w)$ is represented as $\frac{\partial}{\partial w} U(w)$

The goal is to make them appear as $U'(g w - c)$ and $U'(w)$.

There are two approaches:

  • use replacements in the resulting latex string
  • use substitutions in the expression tree

The answer by @dsejas covers string replacements in the latex form.

Here we discuss substitutions in the expression tree.

First we give a ready-made solution. Then we explain how one could explore to get to that solution.

Make a particular derivative display with prime

Here is a function for operator substitution in an expression:

def subs_op(expr, op1, op2):
    r"""
    Return the expression `expr` with operator `op1` replaced by `op2`.
    """
    op = expr.operator()
    if op:
        opp = (op2 if str(op) == str(op1) else op)
        args = expr.operands()
        aargs = (subs_op(a, op1, op2) for a in args)
        return opp(*aargs)
    else:
        return expr

To address the question with this, define the to-be-replaced and the replacement operators, then use subs_op:

sage: DU = U(x).diff().operator()
sage: Uprime = function("U'")
sage: fff = subs_op(ff, DU, Uprime)
sage: fff
-g*(p - 1)*U'(g*w - c) + p*U'(w)
sage: view(fff)

$$ -g $$-g {\left(p - 1\right)} {\rm U'}\left(g w - c\right) + p {\rm U'}\left(w\right) $$U'}\left(w\right)$$

Problem solved, at least in an ad-hoc way.

Make all derivatives display with prime

Here is a use_prime function that walks through an expression and replaces any derivatives of functions of a single variable by "prime" versions (or "double-prime", etc. depending on the order of differentiation).

def use_prime(expr):
    r"""
    Return the expression with one-variable function derivatives "primed".
    """
    op = expr.operator()
    if op:
        args = expr.operands()
        aargs = (use_prime(a) for a in args)
        opp = op
        if (isinstance(op, sage.symbolic.operators.FDerivativeOperator)
                and len(args) == 1):
            name = op.function().name()
            primes = "'" * len(op.parameter_set())
            opp = function(f"{name}{primes}")
        return opp(*aargs)
    else:
        return expr

Example with various types of derivatives:

sage: a, b = (function(f) for f in 'ab')
sage: x, z = SR.var('x, z')
sage: f(x, z) = z * a(x * z) + x * z * diff(b(z), z)
sage: g(x, z) h = diff(f(x, z), x) + diff(f(x, z), z)
sage: h = diff(g(x, z), z, x)
sage: h
x*z^2*D[0, 0](a)(x*z) + z^3*D[0, 0](a)(x*z) + 2*z*D[0](a)(x*z) + z*diff(c(z), z*diff(b(z), z, z) + diff(c(z), diff(b(z), z)
sage: view(h)

$$ x $$x z^{2} \mathrm{D}_{0, 0}\left(a\right)\left(x z\right) + z^{3} \mathrm{D}_{0, 0}\left(a\right)\left(x z\right) {\rm D}_{0, 0}(a)(x z) + 2 \, z \mathrm{D}_{0}\left(a\right)\left(x z\right) {\rm D}_0 (a)(x z) + z frac{partial^{2}}{(partial z)^{2}}c\left(z\right) frac{partial^2}{(partial z)^2}b(z) + frac{partial}{partial z}c\left(z\right) $$z}b(z)$$

sage: hh = use_prime(h)
sage: hh
x*z^2*a''(x*z) + z^3*a''(x*z) + 2*z*a'(x*z) + z*c''(z) + c'(z)
z*b''(z) + b'(z)
sage: view(hh)

$$ x $$x z^{2} {\rm a''}\left(x z\right) + z^{3} {\rm a''}\left(x z\right) + 2 \, z {\rm a'}\left(x z\right) + z {\rm c''}\left(z\right) b''}\left(z\right) + {\rm c'}\left(z\right) $$b'}\left(z\right)$$

Example with the original question:

U = function('U')
EU(x, V(x, y, p) = p*U(x) + (1 - p)*U(y)
w0, w, g, c = SR.var('w0, SR.var('w, g, c')
f = EU(w0, g*w0 V(w, g*w - c, p)
h = diff(f, w0)
w)
show(h)
hh = use_prime(h)
show(hh)

$$ -g {\left(p - 1\right)} \mathrm{D}_{0}\left(U\right)\left(g w_{0} - c\right) $$-g {(p - 1)} \mathrm{D}_0(U)(g w - c) + p frac{partial}{partial w_{0}}U\left(w_{0}\right) $$

$$ -g {\left(p - 1\right)} w}U(w)$$

$$-g {(p - 1)} {\rm U'}\left(g w_{0} - c\right) U'}(g w - c) + p {\rm U'}\left(w_{0}\right) $$

U'}(w)$$

Explore the problem to reach a solution

Having provided an approach above, we now explain the exploration that can lead to such an approach.

Define again

sage: U = function('U')
sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)
sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)

Explore the available methods for expression h with h.<TAB> (ie type h. then hit the TAB key).

Find among the available methods operator and operands:

sage: h.operator(), h.operands()
(<function add_vararg at 0x2007603a0>,
 [-g*(p - 1)*D[0](U)(g*w - c), p*diff(U(w), w)])

Give names to the operands and get one level deeper:

sage: a, b = h.operands()
sage: a.operator(), a.operands()
(<function mul_vararg at 0x2007601f0>, [g, p - 1, D[0](U)(g*w - c), -1])
sage: b.operator(), b.operands()
(<function mul_vararg at 0x2007601f0>, [p, diff(U(w), w)])

Name the operands we care about and explore one level further:

sage: _, _, aa, _ = a.operands()
sage: _, bb = b.operands()

sage: aa
D[0](U)(g*w - c)
sage: bb
diff(U(w), w)

sage: aa.operator(), aa.operands()
(D[0](U), [g*w - c])
sage: bb.operator(), bb.operands()
(D[0](U), [w])

Unfortunately aa.operator() and bb.operator() do not compare equal:

sage: a.operator() == b.operator()
False

So we have to resort to string comparison:

sage: str(a.operator()) == str(b.operator())

Now we try to build this operator in a simpler way:

sage: DUx = U(x).diff()
sage: DUx
diff(U(x), x)
sage: DU = DUx.operator()
sage: DU
D[0](U)

Good, now for the replacement we use a symbolic function whose string representation is U'.

sage: function("U'")
sage: function("U'")
U'

All that is left is to recursively replace one operator by the other by walking the expression tree. See subs_op above.

Now for the more general function, we analyse a little further:

sage: type(DU)
<class 'sage.symbolic.operators.FDerivativeOperator'>
sage: isinstance(DU, sage.symbolic.operators.FDerivativeOperator)
True

and exploring DU. + TAB we find function and parameter_set.

The cases when we want to replace are those where the operator is an FDerivativeOperator and has a single variable. See use_prime above.

Besides exploring using TAB and ?, searching existing questions and tickets can also prove useful, see below.

Inspiration and related material

This answer was inspired by answers to similar questions.

Some queries

In particular see @tmonteil's answer to

and some answers to

Making the D operator more accessible is the object of

Display derivative with prime

Setup

To better focus on the actual question, in this answer we

  • use we use simpler names: names w for and V instead of w0 and EU, V for EU
  • skip the assumption that $p \in [0, 1]$
and skip the assumptions on $p$.

Define U (as a symbolic function U, not as the expression U(x)):

sage: U = function('U')

Define a function V of three variables based on U:

sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)

Note that all names on the left hand side of the assignment operator = get defined by that line: V as a function, and x, y, p as symbolic variables. No need to declare them beforehand.

Define three symbolic variables:variables, an expression f based on V and these variables, and another one, differentiating with respect to w:

sage: w, g, c = SR.var('w, g, c')

Define an expression f based on V and these variables:

sage: f = V(w, g*w - c, p)

Differentiate with respect to w:

sage: h = diff(f, w)
sage: view(h)

View the result:

sage: view(h)

$$-g {\left(p - 1\right)} {(p - 1)} {\rm D}_{0}\left(U\right)\left(g D}_0(U)(g w - c\right) c) + p frac{partial}{partial w}U\left(w\right)$$w}U(w)$$

That has two occurrences of $U'$, represented in two ways:

  • $U'(g w - c)$ is represented as $\mathrm{D}_{0}\left(U\right)\left(g w - c\right)$,
  • $U'(w)$ is represented as $\frac{\partial}{\partial w} U(w)$

The goal is to make them appear which we would like to show as $U'(g w - c)$ and $U'(w)$.

There are two approaches:

  • use replacements in the resulting latex string
  • use substitutions in the expression tree
$U'(w)$ instead of ${\rm D}_0(U)(g w - c)$ and $frac{partial}{partial w} U(w)$.

The answer by @dsejas covers an approach using string replacements in the latex form.representation.

Here we discuss an alternative approach performing substitutions in the expression tree.tree before displaying.

First we give provide two versions of this appraoch, one ad-hoc for a ready-made solution. single derivative replacement, one more general. Then we explain explain how one could could explore to get to that solution.these solutions.

Make a particular derivative display with prime

Here is a function for operator substitution in an expression:

def subs_op(expr, op1, op2):
    r"""
    Return the expression `expr` with operator `op1` replaced by `op2`.
    """
    op = expr.operator()
    if op:
        opp = (op2 if str(op) == str(op1) else op)
        args = expr.operands()
        aargs = (subs_op(a, op1, op2) for a in args)
        return opp(*aargs)
    else:
        return expr

To address the question with this, define the to-be-replaced and the replacement operators, then use subs_op:

sage: DU = U(x).diff().operator()
sage: Uprime = function("U'")
sage: DU, Uprime
(D[0](U), U')
sage: fff = subs_op(ff, DU, Uprime)
sage: fff
-g*(p - 1)*U'(g*w - c) + p*U'(w)
sage: view(fff)

$$-g {\left(p - 1\right)} {\rm U'}\left(g w - c\right) + p {\rm U'}\left(w\right)$$

Problem solved, at least in an ad-hoc way.

Make all derivatives display with prime

Here is a use_prime function that walks through an expression and replaces any derivatives of functions of a single variable by "prime" versions (or "double-prime", etc. depending on the order of differentiation).

def use_prime(expr):
    r"""
    Return the expression with one-variable function derivatives "primed".
    """
    op = expr.operator()
    if op:
        args = expr.operands()
        aargs = (use_prime(a) for a in args)
        opp = op
        if (isinstance(op, sage.symbolic.operators.FDerivativeOperator)
                and len(args) == 1):
            name = op.function().name()
            primes = "'" * len(op.parameter_set())
            opp = function(f"{name}{primes}")
        return opp(*aargs)
    else:
        return expr

Example with various types of derivatives:

sage: a, b = (function(f) for f in 'ab')
sage: x, z = SR.var('x, z')
sage: f(x, z) = z * a(x * z) + x * z * diff(b(z), z)
sage: h = diff(f(x, z), z, x)
sage: h
x*z^2*D[0, 0](a)(x*z) + 2*z*D[0](a)(x*z) + z*diff(b(z), z, z) + diff(b(z), z)
sage: view(h)

$$x z^{2} {\rm D}_{0, 0}(a)(x z) + 2 z {\rm D}_0 (a)(x z) + z frac{partial^2}{(partial z)^2}b(z) + frac{partial}{partial z}b(z)$$

sage: hh = use_prime(h)
sage: hh
x*z^2*a''(x*z) + 2*z*a'(x*z) + z*b''(z) + b'(z)
sage: view(hh)

$$x z^{2} {\rm a''}\left(x z\right) + 2 z {\rm a'}\left(x z\right) + z {\rm b''}\left(z\right) + {\rm b'}\left(z\right)$$

Example with the original question:

U = function('U')
V(x, y, p) = p*U(x) + (1 - p)*U(y)
w, g, c = SR.var('w, g, c')
f = V(w, g*w - c, p)
h = diff(f, w)
show(h)
hh = use_prime(h)
show(hh)

$$-g {(p - 1)} \mathrm{D}_0(U)(g w - c) + p frac{partial}{partial w}U(w)$$

$$-g {(p - 1)} {\rm U'}(g w - c) + p {\rm U'}(w)$$

Explore the problem to reach a solution

Having provided an approach above, we now explain the exploration that can lead to such an approach.

Define again

sage: U = function('U')
sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)
sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)

Explore the available methods for expression h with h.<TAB> (ie type h. then hit the TAB key).

Find among the available methods operator and operands:

sage: h.operator(), h.operands()
(<function add_vararg at 0x2007603a0>,
 [-g*(p - 1)*D[0](U)(g*w - c), p*diff(U(w), w)])

Give names to the operands and get one level deeper:

sage: a, b = h.operands()
sage: a.operator(), a.operands()
(<function mul_vararg at 0x2007601f0>, [g, p - 1, D[0](U)(g*w - c), -1])
sage: b.operator(), b.operands()
(<function mul_vararg at 0x2007601f0>, [p, diff(U(w), w)])

Name the operands we care about and explore one level further:

sage: _, _, aa, _ = a.operands()
sage: _, bb = b.operands()

sage: aa
D[0](U)(g*w - c)
sage: bb
diff(U(w), w)

sage: aa.operator(), aa.operands()
(D[0](U), [g*w - c])
sage: bb.operator(), bb.operands()
(D[0](U), [w])

Unfortunately aa.operator() and bb.operator() do not compare equal:

sage: a.operator() == b.operator()
False

So we have to resort to string comparison:

sage: str(a.operator()) == str(b.operator())

Now we try to build this operator in a simpler way:

sage: DUx = U(x).diff()
sage: DUx
diff(U(x), x)
sage: DU = DUx.operator()
sage: DU
D[0](U)

Good, now for the replacement we use a symbolic function whose string representation is U'.

sage: function("U'")
sage: function("U'")
U'

All that is left is to recursively replace one operator by the other by walking the expression tree. See subs_op above.

Now for the more general function, we analyse a little further:

sage: type(DU)
<class 'sage.symbolic.operators.FDerivativeOperator'>
sage: isinstance(DU, sage.symbolic.operators.FDerivativeOperator)
True

and exploring DU. + TAB we find function and parameter_set.

The cases when we want to replace are those where the operator is an FDerivativeOperator and has a single variable. See use_prime above.

Besides exploring using TAB and ?, searching existing questions and tickets can also prove useful, see below.

Inspiration and related material

This answer was inspired by answers to similar questions.

Some queries

In particular see @tmonteil's answer to

and some answers to

Making the D operator more accessible is the object of

Display derivative with prime

Setup

To better focus on the actual question, in this answer we use simpler names w and V instead of w0 and EU, and skip the assumptions on $p$.

Define U (as a symbolic function U, not as the expression U(x)):

sage: U = function('U')

Define a function V of three variables based on U:

sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)

Note that all names on the left hand side of the assignment operator = get defined by that line: V as a function, and x, y, p as symbolic variables. No need to declare them beforehand.

Define three symbolic variables, an expression f based on V and these variables, and another one, differentiating with respect to w:

sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)
sage: view(h)

$$-g {(p - 1)} {\rm D}_0(U)(g w - c) + p frac{partial}{partial w}U(w)$$frac{∂}{∂ w} U(w)$$

That has two occurrences of $U'$, which we would like to show as $U'(g w - c)$ and $U'(w)$ instead of ${\rm D}_0(U)(g w - c)$ and $frac{partial}{partial w} U(w)$.

The answer by @dsejas covers an approach using string replacements in the latex representation.

Here we discuss an alternative approach performing substitutions in the expression tree before displaying.

First we provide two versions of this appraoch, one ad-hoc for a single derivative replacement, one more general. Then we explain how one could explore to get to these solutions.

Make a particular derivative display with prime

Here is a function for operator substitution in an expression:

def subs_op(expr, op1, op2):
    r"""
    Return the expression `expr` with operator `op1` replaced by `op2`.
    """
    op = expr.operator()
    if op:
        opp = (op2 if str(op) == str(op1) else op)
        args = expr.operands()
        aargs = (subs_op(a, op1, op2) for a in args)
        return opp(*aargs)
    else:
        return expr

To address the question with this, define the to-be-replaced and the replacement operators, then use subs_op:

sage: DU = U(x).diff().operator()
sage: Uprime = function("U'")
sage: DU, Uprime
(D[0](U), U')
sage: fff = subs_op(ff, DU, Uprime)
sage: fff
-g*(p - 1)*U'(g*w - c) + p*U'(w)
sage: view(fff)

$$-g {\left(p - 1\right)} {(p - 1)} {\rm U'}\left(g U'}(g w - c\right) c) + p {\rm U'}\left(w\right)$$U'}(w)$$

Problem solved, at least in an ad-hoc way.

Make all derivatives display with prime

Here is a use_prime function that walks through an expression and replaces any derivatives of functions of a single variable by "prime" versions (or "double-prime", etc. depending on the order of differentiation).

def use_prime(expr):
    r"""
    Return the expression with one-variable function derivatives "primed".
    """
    op = expr.operator()
    if op:
        args = expr.operands()
        aargs = (use_prime(a) for a in args)
        opp = op
        if (isinstance(op, sage.symbolic.operators.FDerivativeOperator)
                and len(args) == 1):
            name = op.function().name()
            primes = "'" * len(op.parameter_set())
            opp = function(f"{name}{primes}")
        return opp(*aargs)
    else:
        return expr

Example with various types of derivatives:

sage: a, b = (function(f) for f in 'ab')
sage: x, z = SR.var('x, z')
sage: f(x, z) = z * a(x * z) + x * z * diff(b(z), z)
sage: h = diff(f(x, z), z, x)
sage: h
x*z^2*D[0, 0](a)(x*z) + 2*z*D[0](a)(x*z) + z*diff(b(z), z, z) + diff(b(z), z)
sage: view(h)

$$x z^{2} z^2 {\rm D}_{0, 0}(a)(x D}_{0,0}(a)(x z) + 2 z {\rm D}_0 (a)(x z) + z frac{partial^2}{(partial frac{∂^2}{(∂ z)^2}b(z) + frac{partial}{partial frac{∂}{∂ z}b(z)$$

sage: hh = use_prime(h)
sage: hh
x*z^2*a''(x*z) + 2*z*a'(x*z) + z*b''(z) + b'(z)
sage: view(hh)

$$x z^{2} z^2 {\rm a''}\left(x z\right) a''}(x z) + 2 z {\rm a'}\left(x z\right) a'}(x z) + z {\rm b''}\left(z\right) b''}(z) + {\rm b'}\left(z\right)$$b'}(z)$$

Example with the original question:

U = function('U')
V(x, y, p) = p*U(x) + (1 - p)*U(y)
w, g, c = SR.var('w, g, c')
f = V(w, g*w - c, p)
h = diff(f, w)
show(h)
hh = use_prime(h)
show(hh)

$$-g {(p - 1)} \mathrm{D}_0(U)(g {\rm D}_0(U)(g w - c) + p frac{partial}{partial frac{∂}{∂ w}U(w)$$

$$-g {(p - 1)} {\rm U'}(g w - c) + p {\rm U'}(w)$$

Explore the problem to reach a solution

Having provided an approach above, we now explain the exploration that can lead to such an approach.

Define again

sage: U = function('U')
sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)
sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)

Explore the available methods for expression h with h.<TAB> (ie type h. then hit the TAB key).

Find among the available methods operator and operands:

sage: h.operator(), h.operands()
(<function add_vararg at 0x2007603a0>,
 [-g*(p - 1)*D[0](U)(g*w - c), p*diff(U(w), w)])

Give names to the operands and get one level deeper:

sage: a, b = h.operands()
sage: a.operator(), a.operands()
(<function mul_vararg at 0x2007601f0>, [g, p - 1, D[0](U)(g*w - c), -1])
sage: b.operator(), b.operands()
(<function mul_vararg at 0x2007601f0>, [p, diff(U(w), w)])

Name the operands we care about and explore one level further:

sage: _, _, aa, _ = a.operands()
sage: _, bb = b.operands()

sage: aa
D[0](U)(g*w - c)
sage: bb
diff(U(w), w)

sage: aa.operator(), aa.operands()
(D[0](U), [g*w - c])
sage: bb.operator(), bb.operands()
(D[0](U), [w])

Unfortunately aa.operator() and bb.operator() do not compare equal:

sage: a.operator() == b.operator()
False

So we have to resort to string comparison:

sage: str(a.operator()) == str(b.operator())

Now we try to build this operator in a simpler way:

sage: DUx = U(x).diff()
sage: DUx
diff(U(x), x)
sage: DU = DUx.operator()
sage: DU
D[0](U)

Good, now for the replacement we use a symbolic function whose string representation is U'.

sage: function("U'")
sage: function("U'")
U'

All that is left is to recursively replace one operator by the other by walking the expression tree. See subs_op above.

Now for the more general function, we analyse a little further:

sage: type(DU)
<class 'sage.symbolic.operators.FDerivativeOperator'>
sage: isinstance(DU, sage.symbolic.operators.FDerivativeOperator)
True

and exploring DU. + TAB we find function and parameter_set.

The cases when we want to replace are those where the operator is an FDerivativeOperator and has a single variable. See use_prime above.

Besides exploring using TAB and ?, searching existing questions and tickets can also prove useful, see below.

Inspiration and related material

This answer was inspired by answers to similar questions.

Some queries

In particular see @tmonteil's answer to

and some answers to

Making the D operator more accessible is the object of

Display derivative with prime

Setup

To better focus on the actual question, in this answer we use simpler names w and V instead of w0 and EU, and skip the assumptions on $p$.

Define U (as a symbolic function U, not as the expression U(x)):

sage: U = function('U')

Define a function V of three variables based on U:

sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)

Note that all names on the left hand side of the assignment operator = get defined by that line: V as a function, and x, y, p as symbolic variables. No need to declare them beforehand.

Define three symbolic variables, an expression f based on V and these variables, and another one, differentiating with respect to w:

sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)
sage: view(h)

$$-g {(p - 1)} {\rm D}_0(U)(g w - c) + p frac{∂}{∂ w} ({∂}/{∂ w}) U(w)$$

That has two occurrences of $U'$, which we would like to show as $U'(g w - c)$ and $U'(w)$ instead of ${\rm D}_0(U)(g w - c)$ and $frac{partial}{partial w} $({∂}/{∂ w}) U(w)$.

The answer by @dsejas covers an approach using string replacements in the latex representation.

Here we discuss an alternative approach performing substitutions in the expression tree before displaying.

First we provide two versions of this appraoch, approach, one ad-hoc for a single derivative replacement, one more general. Then we explain how one could explore to get to these solutions.

Make a particular derivative display with prime

Here is a function for operator substitution in an expression:

def subs_op(expr, op1, op2):
    r"""
    Return the expression `expr` with operator `op1` replaced by `op2`.
    """
    op = expr.operator()
    if op:
        opp = (op2 if str(op) == str(op1) else op)
        args = expr.operands()
        aargs = (subs_op(a, op1, op2) for a in args)
        return opp(*aargs)
    else:
        return expr

To address the question with this, define the to-be-replaced and the replacement operators, then use subs_op:

sage: DU = U(x).diff().operator()
sage: Uprime = function("U'")
sage: DU, Uprime
(D[0](U), U')
sage: fff = subs_op(ff, DU, Uprime)
sage: fff
-g*(p - 1)*U'(g*w - c) + p*U'(w)
sage: view(fff)

$$-g {(p - 1)} {\rm U'}(g w - c) + p {\rm U'}(w)$$

Problem solved, at least in an ad-hoc way.

Make all derivatives display with prime

Here is a use_prime function that walks through an expression and replaces any derivatives of functions of a single variable by "prime" versions (or "double-prime", etc. depending on the order of differentiation).

def use_prime(expr):
    r"""
    Return the expression with one-variable function derivatives "primed".
    """
    op = expr.operator()
    if op:
        args = expr.operands()
        aargs = (use_prime(a) for a in args)
        opp = op
        if (isinstance(op, sage.symbolic.operators.FDerivativeOperator)
                and len(args) == 1):
            name = op.function().name()
            primes = "'" * len(op.parameter_set())
            opp = function(f"{name}{primes}")
        return opp(*aargs)
    else:
        return expr

Example with various types of derivatives:

sage: a, b = (function(f) for f in 'ab')
sage: x, z = SR.var('x, z')
sage: f(x, z) = z * a(x * z) + x * z * diff(b(z), z)
sage: h = diff(f(x, z), z, x)
sage: h
x*z^2*D[0, 0](a)(x*z) + 2*z*D[0](a)(x*z) + z*diff(b(z), z, z) + diff(b(z), z)
sage: view(h)

$$x z^2 {\rm D}_{0,0}(a)(x z) + 2 z {\rm D}_0 (a)(x z) + z frac{∂^2}{(∂ z)^2}b(z) + frac{∂}{∂ z}b(z)$$({∂^2}/{(∂ z)^2})b(z) + ({∂}/{∂ z})b(z)$$

sage: hh = use_prime(h)
sage: hh
x*z^2*a''(x*z) + 2*z*a'(x*z) + z*b''(z) + b'(z)
sage: view(hh)

$$x z^2 {\rm a''}(x z) + 2 z {\rm a'}(x z) + z {\rm b''}(z) + {\rm b'}(z)$$

Example with the original question:

U = function('U')
V(x, y, p) = p*U(x) + (1 - p)*U(y)
w, g, c = SR.var('w, g, c')
f = V(w, g*w - c, p)
h = diff(f, w)
show(h)
hh = use_prime(h)
show(hh)

$$-g {(p - 1)} {\rm D}_0(U)(g w - c) + p frac{∂}{∂ w}U(w)$$({∂}/{∂ z})U(w)$$

$$-g {(p - 1)} {\rm U'}(g w - c) + p {\rm U'}(w)$$

Explore the problem to reach a solution

Having provided an approach above, we now explain the exploration that can lead to such an approach.

Define again

sage: U = function('U')
sage: V(x, y, p) = p*U(x) + (1 - p)*U(y)
sage: w, g, c = SR.var('w, g, c')
sage: f = V(w, g*w - c, p)
sage: h = diff(f, w)

Explore the available methods for expression h with h.<TAB> (ie type h. then hit the TAB key).

Find among the available methods operator and operands:

sage: h.operator(), h.operands()
(<function add_vararg at 0x2007603a0>,
 [-g*(p - 1)*D[0](U)(g*w - c), p*diff(U(w), w)])

Give names to the operands and get one level deeper:

sage: a, b = h.operands()
sage: a.operator(), a.operands()
(<function mul_vararg at 0x2007601f0>, [g, p - 1, D[0](U)(g*w - c), -1])
sage: b.operator(), b.operands()
(<function mul_vararg at 0x2007601f0>, [p, diff(U(w), w)])

Name the operands we care about and explore one level further:

sage: _, _, aa, _ = a.operands()
sage: _, bb = b.operands()

sage: aa
D[0](U)(g*w - c)
sage: bb
diff(U(w), w)

sage: aa.operator(), aa.operands()
(D[0](U), [g*w - c])
sage: bb.operator(), bb.operands()
(D[0](U), [w])

Unfortunately aa.operator() and bb.operator() do not compare equal:

sage: a.operator() == b.operator()
False

So we have to resort to string comparison:

sage: str(a.operator()) == str(b.operator())

Now we try to build this operator in a simpler way:

sage: DUx = U(x).diff()
sage: DUx
diff(U(x), x)
sage: DU = DUx.operator()
sage: DU
D[0](U)

Good, now for the replacement we use a symbolic function whose string representation is U'.

sage: function("U'")
sage: function("U'")
U'

All that is left is to recursively replace one operator by the other by walking the expression tree. See subs_op above.

Now for the more general function, we analyse a little further:

sage: type(DU)
<class 'sage.symbolic.operators.FDerivativeOperator'>
sage: isinstance(DU, sage.symbolic.operators.FDerivativeOperator)
True

and exploring DU. + TAB we find function and parameter_set.

The cases when we want to replace are those where the operator is an FDerivativeOperator and has a single variable. See use_prime above.

Besides exploring using TAB and ?, searching existing questions and tickets can also prove useful, see below.

Inspiration and related material

This answer was inspired by answers to similar questions.

Some queries

In particular see @tmonteil's answer to

and some answers to

Making the D operator more accessible is the object of

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