The D command serves different purposes. It is primarily a differentiation command, like diff (see also comparison), but it is more general than diff in two aspects: it can represent derivatives evaluated at a point and can differentiate procedures.

As a differentiation command, given f, the name of an unknown function of one argument, the call $\mathrm{D}\left(f\right)\left(x\right)$ represents the derivative of f(x) with respect to x. That is, $\mathrm{D}\left(f\right)\left(x\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)$. To switch between these two representations use convert/D and convert/diff.

In the case of partial derivatives, D identifies the differentiation variables by their numerical position. D[i](f) computes the partial derivative of f with respect to its ith argument.  For example,

D[1,1,2](g)(x,y);

${\mathrm{D}}_{1,1,2}\left(g\right)\left(x\,y\right)$

convert((1), diff);

$\frac{{\partial }^3}{\partial x^2\partial y}g\left(x\,y\right)$

Note: It is assumed that partial derivatives commute so that ${\mathrm{D}}_{i,j}\left(g\right)={\mathrm{D}}_{j,i}\left(g\right)$. In the univariate case, ${\mathrm{D}}_1\left(f\right)\left(x\right)$ automatically simplifies to $\mathrm{D}\left(f\right)\left(x\right)$. The zeroth order derivative is represented by D[](g)(x,y) = $\mathrm{diff}\left(g\left(x\,y\right)\,\left[\right]\right)$ = $g\left(x\,y\right)$.

The D command can represent derivatives evaluated at a point. For example, the derivative of $f\left(x\right)$ evaluated at $x=0$ is represented by

D(f)(0);

$\mathrm{D}\left(f\right)\left(0\right)$

The evaluation point can be any algebraic expression, not necessarily a constant. Derivatives evaluated at a point can also be expressed using a composition of eval and diff.

convert((3), diff);

$\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆ\mathrm{t1}}f\left(\mathrm{t1}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆ\mathrm{t1}}f\left(\mathrm{t1}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}$

The D command is appropriate to represent partial derivatives at fixed values of certain functions. For example, the partial derivative of $F\left(t\,v\left(t\right)\right)$ with respect to t at fixed $v\left(t\right)$ is represented by

D[1](F)(t,v(t));

${\mathrm{D}}_1\left(F\right)\left(t\,v\left(t\right)\right)$

A diff representation for this derivative also requires the composition with eval

convert((5), diff);

$\genfrac{}{}{0ex}{}{\left(\frac{\partial }{\partial t}F\left(t\,\mathrm{t1}\right)\right)}{\phantom{\left\{\mathrm{t1}=v\left(t\right)\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\partial }{\partial t}F\left(t\,\mathrm{t1}\right)\right)}}{\left\{\mathrm{t1}=v\left(t\right)\right\}}$

In the same manner, D can be used to represent the derivative of $F\left(t\,v\left(t\right)\right)$ "with respect to $v\left(t\right)$", a function.

D[2](F)(t,v(t));

${\mathrm{D}}_2\left(F\right)\left(t\,v\left(t\right)\right)$

convert((7), diff);

$\genfrac{}{}{0ex}{}{\left(\frac{\partial }{\partial \mathrm{t1}}F\left(t\,\mathrm{t1}\right)\right)}{\phantom{\left\{\mathrm{t1}=v\left(t\right)\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\partial }{\partial \mathrm{t1}}F\left(t\,\mathrm{t1}\right)\right)}}{\left\{\mathrm{t1}=v\left(t\right)\right\}}$

The D command can differentiate procedures, typically used to represent mathematical mappings, as well as arbitrary algebraic expressions understood as mappings, in the sense that in Maple algebraic expressions are appliable. This functionality is unique to D; it cannot be emulated using the diff command.

The derivative of a procedure or mapping of n variables is another mapping of the same number of variables. For example, D(sin) returns cos. In the general case, $\mathrm{D}\left(f\right)$ represents the mapping

D(f) = (q -> eval(diff(f(t),t), t=q));

$\mathrm{D}\left(f\right)=\left(q\mapsto \genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆt}f\left(t\right)\right)}{\phantom{t=q}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆt}f\left(t\right)\right)}}{t=q}\right)$

so that when $\mathrm{D}\left(f\right)$ is applied to some variable, say x, or some expression, you have

(9)(x);

$\mathrm{D}\left(f\right)\left(x\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)$

(9)(x+1);

$\mathrm{D}\left(f\right)\left(x+1\right)=\mathrm{D}\left(f\right)\left(x+1\right)$

When differentiating an arbitrary mapping represented by an algebraic expression, say F, the expression can contain constants, known function names such as sin, exp (not the applied function exp(x)), unknown function names such as f, g, arrow operators such as $x\to x^2$, and the arithmetic and functional operators. On such expressions, D recursively applies the rules

Note: The composition of functions, say f and g, should be entered using the notation $f@g$, not $f\left(g\right)$.  To specify that a symbol, say g, is a constant and not a function name, use the assuming command, like in D(f*g) assuming g::constant; see also assume.

When F is a name to which a procedure of n arguments is assigned, D can compute the mappings representing the partial derivatives of F (restrictions apply). The procedure F can have assignments to local variables, if statements, for loops, while loops, etc.

It is an error to specify a call to the derivative of a function with respect to the k-th variable and apply it to fewer than k variables. For instance, ${\mathrm{D}}_2\left(f\right)\left(x\right)$ yields an error.

$\mathrm{D}\left(\ln \right)\left(x\right)=\mathrm{diff}\left(\ln \left(x\right)\,x\right)$

$\frac{1}{x}=\frac{1}{x}$

$\mathrm{D}\left(y\right)\left(x\right)$

$\mathrm{D}\left(y\right)\left(x\right)$

$\mathrm{convert}\left(\,\mathrm{diff}\right)$

$\frac{\ⅆ}{\ⅆx}y\left(x\right)$

$\mathrm{convert}\left(\,\mathrm{D}\right)$

$\mathrm{D}\left(y\right)\left(x\right)$

$\mathrm{D}\left(\sin \right)$

$\cos$

${\mathrm{D}}^{\left(n\right)}\left(\sin \right)$

$z\mapsto \sin \left(z+\frac{n\cdot \mathrm{\pi }}{2}\right)$

$\mathrm{D}\left[\mathrm{`\$`}\left(1\,n\right)\right]\left(\sin \right)$

$z\mapsto \sin \left(z+\frac{n\cdot \mathrm{\pi }}{2}\right)$

$\mathrm{D}\left(f\right)$

$\mathrm{D}\left(f\right)$

$\mathrm{D}\left(\mathrm{D}\left(f\right)\right)$

${\mathrm{D}}^{\left(2\right)}\left(f\right)$

${\mathrm{D}}^{\left(2\right)}\left(f\right)$

${\mathrm{D}}^{\left(2\right)}\left(f\right)$

$\mathrm{D}\left[\mathrm{`\$`}\left(1\,n\right),\mathrm{`\$`}\left(2\,m\right)\right]\left(g\right)$

${\mathrm{D}}_{\underset{\text{n times}}{\underbrace{1,\mathrm{...},1}},\underset{\text{m times}}{\underbrace{2,\mathrm{...},2}}}\left(g\right)$

$\mathrm{D}\left[i,j\right]\left(g\right)-\mathrm{D}\left[j,i\right]\left(g\right)$

$0$

$\mathrm{D}\left[1\right]\left(\mathrm{D}\left[2,1\right]\left(g\right)\right)$

${\mathrm{D}}_{1,1,2}\left(g\right)$

$\mathrm{D}\left(\arctan \right)$

$z\mapsto \frac{1}{z^2+1}$

$\mathrm{D}\left[1\right]\left(\arctan \right)$

$\left(y\,x\right)\mapsto \frac{1}{x\cdot \left(1+\frac{y^2}{x^2}\right)}$

$\mathrm{D}\left[2\right]\left(\mathrm{{\rm Z}}\right)$

${\mathrm{D}}_2\left(\mathrm{{\rm Z}}\right)$

$\left(x\,y\right)\ne \left(x\,y\,z\right)$

${\mathrm{\zeta }}^{\left(x+1\right)}\left(y\right)\ne {\mathrm{\zeta }}^{\left(x+1\right)}\left(y\,z\right)$

$\mathrm{D}\left(\exp +{\cos }^2+\mathrm{\pi }+\tan \right)$

$-2\sin \cos +{\tan }^2+\exp +1$

$\mathrm{D}\left(2f\sin \right)$

$2\mathrm{D}\left(f\right)\sin +2f\cos$

$\mathrm{D}\left(2f\sin \right)\mathbf{assuming}f::\mathrm{constant}$

$2f\cos$

$f≔\left(x\,y\right)\mapsto \exp \left(y\cdot x\right)$

$f≔\left(x\,y\right)\mapsto {\ⅇ}^{y\cdot x}$

$\mathrm{D}\left[\right]\left(f\right)$

$f$

$\mathrm{D}\left[1\right]\left(f\right)$

$\left(x\,y\right)\mapsto y\cdot {\ⅇ}^{y\cdot x}$

$\mathrm{D}\left[2\right]\left(f\right)$

$\left(x\,y\right)\mapsto x\cdot {\ⅇ}^{y\cdot x}$

$H≔\left(p\,x\right)\mapsto \frac{p^2}{2\cdot m}+\frac{k\cdot x^2}{2}$

$H≔\left(p\,x\right)\mapsto \frac{p^2}{2\cdot m}+\frac{k\cdot x^2}{2}$

$\mathrm{eq}\left[1\right]≔\mathrm{diff}\left(x\left(t\right)\,t\right)=\mathrm{D}\left[1\right]\left(H\right)\left(p\left(t\right)\,x\left(t\right)\right)$

${\mathrm{eq}}_1≔\frac{\ⅆ}{\ⅆt}x\left(t\right)=\frac{p\left(t\right)}{m}$

$\mathrm{eq}\left[2\right]≔\mathrm{diff}\left(p\left(t\right)\,t\right)=-\mathrm{D}\left[2\right]\left(H\right)\left(p\left(t\right)\,x\left(t\right)\right)$

${\mathrm{eq}}_2≔\frac{\ⅆ}{\ⅆt}p\left(t\right)=-kx\left(t\right)$

$\mathrm{simplify}\left(\mathrm{eq}\left[2\right]\,\left\{\mathrm{eq}\left[1\right]\right\}\,\left\{p\left(t\right)\right\}\right)$

$\left(\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)\right)m=-kx\left(t\right)$

$\mathrm{diff}\left(E\left(t\right)=H\left(p\left(t\right)\,x\left(t\right)\right)\,t\right)$

$\frac{\ⅆ}{\ⅆt}E\left(t\right)=\frac{p\left(t\right)\left(\frac{\ⅆ}{\ⅆt}p\left(t\right)\right)}{m}+kx\left(t\right)\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)$

$\mathrm{eval}\left(\,\left[\mathrm{eq}\left[1\right]\,\mathrm{eq}\left[2\right]\right]\right)$

$\frac{\ⅆ}{\ⅆt}E\left(t\right)=0$