Differentiation Rules for Calculus1 - Maple Programming Help

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Differentiation Rules for Calculus1

Rules

 • See Student[Calculus1] for a general introduction to the Calculus1 subpackage of the Student package.
 • See SingleStepOverview for an introduction to the step-by-step (or single-step) functionality of the Calculus1 package.
 • The following table lists the built-in rules for differentiation that do not take parameters.  These rules can be passed as the index to Rule or as a rule argument to Understand.

 Rule Alternate Names Description chain $\left(f\left(g\left(x\right)\right)\right)\prime =f\prime \left(g\left(x\right)\right)g\prime \left(x\right)$ constant $c\prime =0$ constantmultiple $\mathrm{c*}$ $\left(cf\right)\prime =cf\prime$ difference $\mathrm{-}$ $\left(f-g\right)\prime =f\prime -g\prime$ identity $\mathrm{^}$ $x\prime =1$ int $\mathrm{Int}$ $\left({{\int }}_{c}^{x}f\left(t\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}t\right)\text{'}=f\left(x\right)$ power $\mathrm{^}$ $\left({x}^{n}\right)\prime =n{x}^{n-1}$ product $\mathrm{*}$ $\left(fg\right)\prime =f\prime g+fg\prime$ quotient $\mathrm{/}$ $\left(\frac{f}{g}\right)\prime =\frac{gf\prime -fg\prime }{{g}^{2}}$ sum $\mathrm{+}$ $\left(f+g\right)\prime =f\prime +g\prime$

 The name of any univariate function can also be used as a rule argument to the Rule command.  The name of any univariate function recognized by Maple, for example, sin, can be passed as a rule argument to the Understand command (where recognized means that it is of type mathfunc).
 • There is one differentiation rule which requires a parameter: rewrite.  This rule can be used as the index to a call to Rule, but cannot be given as a rule argument to Understand.  This rule is used to change the form of the expression being differentiated.  It has the general form:

 [rewrite, $\mathrm{f1}\left(x\right)=\mathrm{g1}\left(x\right)$, $\mathrm{f2}\left(x\right)=\mathrm{g2}\left(x\right)$, ...]

 The effect of applying the rewrite rule is to perform each substitution listed as a parameter to the rule, where occurrences of the left-hand side of each substitution are replaced by the corresponding right-hand side.
 The main application of this rule is to rewrite an expression of the form ${f\left(x\right)}^{g\left(x\right)}$, where the exponent (at least) depends on the differentiation variable, as an exponential.  The rule would thus be given as:

 [rewrite, ${f\left(x\right)}^{g\left(x\right)}={ⅇ}^{g\left(x\right)\mathrm{ln}\left(f\left(x\right)\right)}$ ]

 Note: The Rule routine does not attempt to validate the rewrite rules you provide.

Examples

 > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{Calculus1}\right):$
 > $\mathrm{infolevel}\left[\mathrm{Student}\left[\mathrm{Calculus1}\right]\right]≔1:$
 > $\mathrm{Rule}\left[\mathrm{*}\right]\left(\mathrm{Diff}\left({x}^{2}\mathrm{sin}\left({x}^{2}\right),x\right)\right)$
 Creating problem #1
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{}{\mathrm{sin}}{}\left({{x}}^{{2}}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){}{\mathrm{sin}}{}\left({{x}}^{{2}}\right){+}{{x}}^{{2}}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{sin}}{}\left({{x}}^{{2}}\right)$ (1)
 > $\mathrm{Rule}\left[\mathrm{chain}\right]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{}{\mathrm{sin}}{}\left({{x}}^{{2}}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){}{\mathrm{sin}}{}\left({{x}}^{{2}}\right){+}{{x}}^{{2}}{}\left(\genfrac{}{}{0}{}{\frac{{ⅆ}}{{ⅆ}{\mathrm{_X0}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{sin}{}\left({\mathrm{_X0}}\right)}{\phantom{{\mathrm{_X0}}{=}{{x}}^{{2}}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{{ⅆ}}{{ⅆ}{\mathrm{_X0}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{sin}{}\left({\mathrm{_X0}}\right)}}{{\mathrm{_X0}}{=}{{x}}^{{2}}}\right){}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right)$ (2)
 > $\mathrm{Rule}\left[\mathrm{sin}\right]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{}{\mathrm{sin}}{}\left({{x}}^{{2}}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){}{\mathrm{sin}}{}\left({{x}}^{{2}}\right){+}{{x}}^{{2}}{}{\mathrm{cos}}{}\left({{x}}^{{2}}\right){}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right)$ (3)

If the operation type is ambiguous, Maple returns an error

 > $\mathrm{Rule}\left[\mathrm{sum}\right]\left(\mathrm{Diff}\left({x}^{2}+\mathrm{Int}\left(\mathrm{cos}\left(t\right),t=0..x\right),x\right)\right)$
 > $\mathrm{Rule}\left[\mathrm{sum}\right]\left(\mathrm{Diff}\left({x}^{2}+\mathrm{Int}\left(\mathrm{cos}\left(t\right),t=0..x\right),x\right),\mathrm{diff}\right)$
 Creating problem #2
 $\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{+}{{\int }}_{{0}}^{{x}}{\mathrm{cos}}{}\left({t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){+}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{\int }}_{{0}}^{{x}}{\mathrm{cos}}{}\left({t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}\right)$ (4)
 > $\mathrm{Rule}\left[\mathrm{int}\right]\left(\right)$
 $\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{+}{{\int }}_{{0}}^{{x}}{\mathrm{cos}}{}\left({t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){+}{\mathrm{cos}}{}\left({x}\right)$ (5)
 > $\mathrm{Rule}\left[\mathrm{^}\right]\left(\mathrm{Diff}\left(\mathrm{exp}\left(x\right),x\right)\right)$
 Creating problem #3 Rule [power] does not apply
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{exp}{}\left({x}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{exp}{}\left({x}\right)$ (6)
 > $\mathrm{Rule}\left[\mathrm{rewrite},{x}^{\mathrm{sin}\left(x\right)}=\mathrm{exp}\left(\mathrm{sin}\left(x\right)\mathrm{ln}\left(x\right)\right)\right]\left(\mathrm{Diff}\left({x}^{\mathrm{sin}\left(x\right)},x\right)\right)$
 Creating problem #4
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{\mathrm{sin}}{}\left({x}\right)}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{{ⅇ}}^{{\mathrm{sin}}{}\left({x}\right){}{\mathrm{ln}}{}\left({x}\right)}$ (7)

This example illustrates how to handle an unknown univariate function.

 > $\mathrm{Rule}\left[\mathrm{*}\right]\left(\mathrm{Diff}\left(rf\left(r\right),r\right)\right)$
 Creating problem #5
 $\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({r}{}{f}{}\left({r}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{r}{}{f}{}\left({r}\right){+}{r}{}\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}\right)$ (8)
 > $\mathrm{Rule}\left[f\right]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({r}{}{f}{}\left({r}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{r}{}{f}{}\left({r}\right){+}{r}{}\left(\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}\right)\right)$ (9)
 > $\mathrm{Rule}\left[\mathrm{identity}\right]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({r}{}{f}{}\left({r}\right)\right){=}{f}{}\left({r}\right){+}{r}{}\left(\frac{{ⅆ}}{{ⅆ}{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}\right)\right)$ (10)
 > $\mathrm{ShowIncomplete}\left(\right)$
 The current problem is complete