Student[Calculus1]
ShowSolution
show all the steps in the solution of a specified problem
Calling Sequence
Parameters
Description
Examples
Compatibility
ShowSolution(p, opts)
p
-
(optional) posint or a calculus1 problem; the problem to solve
opts
(optional) options of the form keyword=value, where keyword is one of maxsteps, searchoptimal, showrules, output, displaystyle.
The ShowSolution command is used to show the solution steps for a Calculus1 problem, that is, a limit, differentiation or integration problem such as can be expected to be encountered in a single-variable calculus course.
For solution steps to an implicit differentiation problem, see ImplicitDiffSolution .
If p is omitted, the current (most recently referenced) problem is solved. Otherwise, the problem referenced by p is solved. A problem can be referenced either by its problem number (see GetProblem and ShowIncomplete ) or by the problem itself, for example via a label .
These options can be used to control how the problem is solved:
maxsteps = posint (default: 25)
This puts a limit on the number of rules which can be applied to solve a problem.
simplify = truefalse (default: deduce)
By setting simplify=true, the simplify command is applied to the function inside the integral, limit, or derivative before proceeding to the given calculus operation. The default behavior is conservative; it uses the simplified expression only when it is much simpler (in terms of size). Setting simplify=false avoids simplification, which is useful when a specific rule is intended to be applied, and cannot be used on the simplified version of the same input.
searchoptimal = truefalse (default: true)
The default behavior is to try to determine the shortest solution sequence. If this option is given as searchoptimal = false, the first solution discovered is displayed.
These options can be used to control how the problem is displayed:
showrules = truefalse (default: true)
Normally, the rule applied at each step of the solution is displayed; if this option is given as showrules=false, the rules are not shown, and the displaystyle is brief.
output = canvas,script,record,list,print,printf,typeset,link (default: typeset)
The output options are described in Student:-Basics:-OutputStepsRecord
displaystyle= columns,compact,linear,brief (default: linear)
The displaystyle options are described in Student:-Basics:-OutputStepsRecord . Setting displaystyle = brief shows the steps in a compact form with one or two words at each step to describe the rule being applied.
If you set infolevel[Student] := 1 or infolevel[Student[Calculus1]] := 1 (see infolevel ), Maple may display some additional, useful information about the state of the problem and its solution.
with⁡StudentCalculus1:
Diff⁡x2⁢sin⁡x,x
ⅆⅆxx2⁢sin⁡x
ShowSolution⁡
Differentiation Stepsⅆⅆxx2⁢sin⁡x▫1. Apply theproductrule◦Recall the definition of theproductruleⅆⅆxf⁡x⁢g⁡x=ⅆⅆxf⁡x⁢g⁡x+f⁡x⁢ⅆⅆxg⁡xf⁡x=x2g⁡x=sin⁡xThis gives:ⅆⅆxx2⁢sin⁡x+x2⁢ⅆⅆxsin⁡x▫2. Apply thepowerrule to the termⅆⅆxx2◦Recall the definition of thepowerruleⅆⅆxxn=n⁢xn−1◦This means:ⅆⅆxx2=2⋅x1◦So,ⅆⅆxx2=2⋅xWe can rewrite the derivative as:2⋅x⋅sin⁡x+x2⁢ⅆⅆxsin⁡x▫3. Evaluate the derivative ofsin(x)◦Recall the definition of thesinruleⅆⅆxsin⁡x=cos⁡xThis gives:2⁢x⁢sin⁡x+x2⁢cos⁡x
Int⁡sin⁡x2,x
∫sin⁡x2ⅆx
Hint⁡
rewrite,sin⁡x2=12−cos⁡2⁢x2
Rule⁡
∫sin⁡x2ⅆx=∫12−cos⁡2⁢x2ⅆx
Integration Steps∫sin⁡x2ⅆx▫1. Rewrite◦Equivalent expressionsin⁡x2=12−cos⁡2⁢x2This gives:∫12−cos⁡2⁢x2ⅆx▫2. Apply thesumrule◦Recall the definition of thesumrule∫f⁡x+g⁡xⅆx=∫f⁡xⅆx+∫g⁡xⅆxf⁡x=12g⁡x=−cos⁡2⁢x2This gives:∫12ⅆx+∫−cos⁡2⁢x2ⅆx▫3. Apply theconstantrule to the term∫12ⅆx◦Recall the definition of theconstantrule∫Cⅆx=C⁢x◦This means:∫12ⅆx=x2We can now rewrite the integral as:x2+∫−cos⁡2⁢x2ⅆx▫4. Apply theconstant multiplerule to the term∫−cos⁡2⁢x2ⅆx◦Recall the definition of theconstant multiplerule∫C⁢f⁡xⅆx=C⁢∫f⁡xⅆx◦This means:∫−cos⁡2⁢x2ⅆx=−∫cos⁡2⁢xⅆx2We can rewrite the integral as:x2+−∫cos⁡2⁢xⅆx2▫5. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=2⁢x◦Differentiate both sidesdu=2⁢dx◦Isolate equation fordxdx=du2◦Substitute the values for u and dx back into the original∫cos⁡2⁢xⅆx=∫cos⁡u2ⅆuThis gives:x2−∫cos⁡u2ⅆu2▫6. Apply theconstant multiplerule to the term∫cos⁡u2ⅆu◦Recall the definition of theconstant multiplerule∫C⁢f⁡uⅆu=C⁢∫f⁡uⅆu◦This means:∫cos⁡u2ⅆu=∫cos⁡uⅆu2We can rewrite the integral as:x2+−∫cos⁡uⅆu22▫7. Evaluate the integral ofcos(u)◦Recall the definition of thecosrule∫cos⁡uⅆu=sin⁡uThis gives:x2−14⋅sin⁡u▫8. Revert change of variable◦Variable we defined in step5u=2⁢xThis gives:x2−14⋅sin⁡2⁢x•Add constant of integrationx2+−sin⁡2⁢x4+C
Understand⁡Int,`c*`,revert
Int=constantmultiple,revert
ShowSolution⁡Int⁡14⁢x2+4⁢x+1,x
Integration Steps∫14⁢x2+4⁢x+1ⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=x+12◦Differentiate both sidesdu=dx◦Isolate equation fordxdx=du◦Substitute the values for u and dx back into the original∫14⁢x2+4⁢x+1ⅆx=∫1u2ⅆu4This gives:∫1u2ⅆu4▫2. Apply thepowerrule to the term∫1u2ⅆu◦Recall the definition of thepowerrule, for n≠-1∫unⅆu=un+1n+1◦This means:∫1u2ⅆu=u−2+1−2+1◦So,∫1u2ⅆu=−1uWe can rewrite the integral as:−12⁢2⁢x+1•Add constant of integration−12⁢2⁢x+1+C
The output and displaystyle options were introduced in Maple 2021.
For more information on Maple 2021 changes, see Updates in Maple 2021 .
The Student:-Calculus1:-ShowSolution command was updated in Maple 2023.
The simplify option was introduced in Maple 2023.
For more information on Maple 2023 changes, see Updates in Maple 2023 .
See Also
Calculus1
GetProblem
ImplicitDiffSolution
infolevel
label
ShowIncomplete
ShowSteps
Download Help Document
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