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Student[Calculus1]

 Hint
 ask for a hint for a Calculus1 problem

 Calling Sequence Hint(expr, opn)

Parameters

 expr - algebraic or algebraic equation; specify the problem opn - (optional) name; specify problem type: one of Diff, Int, and Limit

Description

 • The Hint(expr) command returns a hint for the problem matching expr.  If no problem in the Calculus1 table of problems for the current session matches expr, a new problem is created for this expression.  The parameter expr must be a(n):
 1 New problem (typically an expression of the form Diff(...), Int(...), or Limit(...))
 2 Output from a previous invocation of Rule
 3 Output from a call to GetProblem
 4 Subproblem label, as displayed by a call to ShowIncomplete (Note: Treat subproblem labels as temporary objects because the application of a rule to a problem can change the underlying problem representation, and hence the subproblem labels.  It is recommended that you call ShowIncomplete to verify the value of a label before passing it to Hint.)
 • This command can be applied interactively by using the DiffTutor, IntTutor, or LimitTutor.
 • Use the Hint(expr, opn) command if expr does not match a problem in the Calculus1 table of problems for the current session and expr includes more than one Calculus1 operation, for example, $\frac{{ⅆ}}{{ⅆ}x}\left(x+{∫}\left(f\left(x\right)+x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x\right)$.  The opn parameter specifies the operation type. If expr matches a problem in the Calculus1 table of problems for the current session, the opn parameter is ignored (that is, you cannot change the operation type of an existing problem).
 • The output from a call to Hint is a list containing a rule, or an expression sequence of rules, if more than one rule can be applied to the problem.  If Hint is unable to find a rule to suggest, it returns the empty list, $\left[\right]$.
 • The output from Hint can be used directly as the index (specifying the rule to apply) to Rule.  That is, if $\mathrm{expr}$ is valid input to Hint, the construct Rule[Hint(expr)](expr) is also valid.  In the case that Hint returns multiple suggestions, this construct applies the first hint to Rule. You can, of course, request that the second, for example, suggestion be applied by using a construct such as Hint(expr); Rule[%[2]](expr);.
 • It is recommended that you use the Maple userinfo system, by setting infolevel[Student] := 1 or infolevel[Student[Calculus1]] := 1. This displays informative messages in the form of userinfo output from Calculus1 routines as you solve complicated problems.

Examples

 > $\mathrm{with}\left(\mathrm{Student}[\mathrm{Calculus1}]\right):$
 > ${\mathrm{infolevel}}_{\mathrm{Student}[\mathrm{Calculus1}]}≔1:$
 > $\frac{{ⅆ}}{{ⅆ}x}\left({x}^{2}+x\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}{}\left({{x}}^{{2}}{+}{x}\right)$ (1)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #1
 $\left[{\mathrm{sum}}\right]$ (2)
 > $\mathrm{Rule}[]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}{}\left({{x}}^{{2}}{+}{x}\right){=}\frac{{ⅆ}}{{ⅆ}{x}}{}\left({{x}}^{{2}}\right){+}\frac{{ⅆ}}{{ⅆ}{x}}{}{x}$ (3)
 > ${∫}x\mathrm{sin}\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x$
 ${∫}{x}{}{\mathrm{sin}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}$ (4)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #2
 $\left[{\mathrm{parts}}{,}{x}{,}{-}{\mathrm{cos}}{}\left({x}\right)\right]$ (5)
 > $\mathrm{Rule}[]\left(\right)$
 ${∫}{x}{}{\mathrm{sin}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{-}{x}{}{\mathrm{cos}}{}\left({x}\right){-}\left({∫}\left({-}{\mathrm{cos}}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)$ (6)

If the first reference to a problem is as the expr argument to Hint, you must use GetProblem to retrieve the problem to pass to Rule.

 > $\mathrm{Hint}\left(\underset{x→{0}^{+}}{{lim}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}x\mathrm{ln}\left(x\right)\right)$
 Creating problem #3
 $\left[{\mathrm{lhopital}}{,}{\mathrm{ln}}{}\left({x}\right)\right]{,}\left[{\mathrm{lhopital}}{,}{x}\right]$ (7)
 > $\mathrm{Rule}[]\left(\mathrm{GetProblem}\left(\mathrm{internal}\right)\right)$
 $\underset{{x}{→}{{0}}^{{+}}}{{lim}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}{}{\mathrm{ln}}{}\left({x}\right){=}\underset{{x}{→}{{0}}^{{+}}}{{lim}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({-}{x}\right)$ (8)
 > ${∫}\frac{t}{\left(t+1\right)\sqrt{{t}^{2}+t+1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}t$
 ${∫}\frac{{t}}{\left({t}{+}{1}\right){}\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (9)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #4 Hints: 1. Complete the square and make a change of variable. 2. Convert the rational part of the integrand to partial fractions and split integrand into a sum of simpler expressions.
 $\left[{\mathrm{change}}{,}{u}{=}{t}{+}\frac{{1}}{{2}}{,}{u}\right]{,}\left[{\mathrm{rewrite}}{,}\frac{{t}}{\left({t}{+}{1}\right){}\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}{=}\frac{{1}}{\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}{-}\frac{{1}}{\left({t}{+}{1}\right){}\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}\right]$ (10)
 > $\mathrm{Rule}[[2]]\left(\right)$
 ${∫}\frac{{t}}{\left({t}{+}{1}\right){}\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{∫}\left(\frac{{1}}{\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}{-}\frac{{1}}{\left({t}{+}{1}\right){}\sqrt{{{t}}^{{2}}{+}{t}{+}{1}}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (11)

If the operation type is ambiguous, Maple returns an error

 > $\mathrm{Hint}\left({∫}\left(x+\frac{{ⅆ}}{{ⅆ}x}\left({x}^{2}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x\right)$
 > $\mathrm{Hint}\left({∫}\left(x+\frac{{ⅆ}}{{ⅆ}x}\left({x}^{2}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x,\mathrm{Diff}\right)$
 Creating problem #5
 $\left[{\mathrm{power}}\right]$ (12)