Revisions to Can we decide whenever a function is the derivate of another function in this Language?

Notice removed Authoritative reference needed by CoffeDeveloper

occurred Nov 26, 2016 at 2:00

Bounty Ended with Emil Jeřábek's answer chosen by CoffeDeveloper

occurred Nov 26, 2016 at 2:00

minor clarifications

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edited Nov 24, 2016 at 14:59

CoffeDeveloper

55
9

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

P.P.S: Use of absolute value is admitted because of following observation:

$$ |f(x)| = sqrt(f(x)*f(x))$$

So you are just using a short form, but you are not taking positive values of a function, you are just computing the Squareroot of its Square in this language. Nor integrals are admitted in this language.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

P.P.S: Use of absolute value is admitted because of following observation:

$$ |f(x)| = sqrt(f(x)*f(x))$$

So you are just using a short form, but you are not taking positive values of a function, you are just computing the Squareroot of its Square.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

P.P.S: Use of absolute value is admitted because of following observation:

$$ |f(x)| = sqrt(f(x)*f(x))$$

So you are just using a short form, but you are not taking positive values of a function, you are just computing the Squareroot of its Square in this language. Nor integrals are admitted in this language.

added 261 characters in body

Source Link

edited Nov 24, 2016 at 14:48

CoffeDeveloper

55
9

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm thatif there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

P.P.S: Use of absolute value is admitted because of following observation:

$$ |f(x)| = sqrt(f(x)*f(x))$$

So you are just using a short form, but you are not taking positive values of a function, you are just computing the Squareroot of its Square.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

P.P.S: Use of absolute value is admitted because of following observation:

$$ |f(x)| = sqrt(f(x)*f(x))$$

So you are just using a short form, but you are not taking positive values of a function, you are just computing the Squareroot of its Square.

fixed power stuff to have only real exponent

Source Link

edited Nov 24, 2016 at 14:37

CoffeDeveloper

55
9

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x),g(x))$$pow( f(x), k)$ is a EXP if $g(x),f(x)$ are$f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$

if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x),g(x))$ is a EXP if $g(x),f(x)$ are EXP
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

Our EXP functions are made in the following way:

Any constant $ \in \Bbb R$ is a EXP
$X \in \Bbb R$ is a EXP
$sin( g(x))$, $cos( g(x))$ are in EXP if $g(x)$ is a EXP
$tan( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \neq \frac\pi 2 + k\pi, k \in \Bbb Z $
$sqrt( g(x))$ is a EXP if $g(x)$ is a EXP and $g(x) \gt 0$
$pow( f(x), k)$ is a EXP if $f(x)$ is a EXP and $k \in \Bbb R$
$exp(g(x))$ is a EXP if $g(x)$ is a EXP
$ln(f(x))$ is a EXP if $f(x)$ is a EXP and $f(x) \gt 0$
$f(x)+g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)-g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)*g(x)$ is a EXP if $g(x),f(x)$ are EXP
$f(x)/g(x)$ is a EXP if $g(x),f(x)$ are EXP and $g(x) \neq 0$
if $g(x)$ is a EXP then $g'(x)$ is a EXP (derivative)

What I want to know, is if there's an algorithm that, given $f(x) \in EXP$ can say:

Yes: iff there exist a $g(x)$ such that $g'(x) = f(x)$ and $g(x) \in EXP$
No: otherwise
Always halt

Of course since we have derivates, also the following derivation rules still applies:

if $f(x) = c$ where $c \in \Bbb R$, then $f'(x) = 0$
if $f(x) = x$ where $x \in \Bbb R$, then $f'(x) = 1$
if $f(x) = sin(g(x))$ then $f'(x) = cos(f(x))*g'(x)$
if $f(x) = cos(g(x))$ then $f'(x) = -sin(f(x))*g'(x)$
if $f(x) = tan(g(x))$ then $f'(x) = \frac{g'(x)}{cos^2(g(x))}$
if $f(x) = sqrt(g(x))$ then $f'(x) = (1/2)*pow(g(x), -1/2)*g'(x)$
if $f(x) = exp(g(x))$ then $f'(x) = exp(g(x))*g'(x)$

if $f(x) = pow(g(x),k)$ and $k \in \Bbb R$ then $f'(x) = k*pow(g(x), k-1)*g'(x)$
if $f(x) = ln(g(x))$ then $f'(x) = \frac{g'(x)}{g(x)}$
if $f(x) = g(x)+h(x)$ then $f'(x) = g'(x)+h'(x)$
if $f(x) = g(x)-h(x)$ then $f'(x) = g'(x)-h'(x)$
if $f(x) = g(x)*h(x)$ then $f'(x) = g'(x)*h(x) +g(x)*h'(x)$
if $f(x) = g(x)/h(x)$ then $f'(x) = \frac{g'(x)*h(x) -g(x)*h'(x)}{h(x)^2}$

P.S. I did not repeated that denominator of a fraction should be different by zero because that is already covered by the first set of rules.

Rollback to Revision 9

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edited Nov 24, 2016 at 14:33

CoffeDeveloper

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removed pow function which is not trivial to derivate nor is taught in high-school

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edited Nov 24, 2016 at 14:32

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more clear and added a punctualization

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edited Nov 24, 2016 at 14:20

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Notice added Authoritative reference needed by CoffeDeveloper

occurred Nov 24, 2016 at 14:16

Bounty Started worth 50 reputation by CoffeDeveloper

occurred Nov 24, 2016 at 14:16

added subset of derivation rules for my language and fixed a few points in the bullet list

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edited Nov 24, 2016 at 14:13

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derivating sqrt in 0 may be a problem

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edited Nov 23, 2016 at 9:58

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added 18 characters in body

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edited Nov 23, 2016 at 9:53

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Made it more formal, and removed a statement that would made the set decidable by definition

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edited Nov 23, 2016 at 9:47

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added 209 characters in body

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edited Nov 22, 2016 at 16:52

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added 255 characters in body

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edited Nov 22, 2016 at 16:32

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added 262 characters in body

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edited Nov 22, 2016 at 14:43

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asked Nov 22, 2016 at 12:32

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