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FunctionAdvisor/differentiation_rule - return the differentiation rule of a given mathematical function
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Calling Sequence
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FunctionAdvisor(differentiation_rule, math_function)
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Parameters
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differentiation_rule
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literal name; 'differentiation_rule'
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math_function
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Maple name of mathematical function
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Description
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The FunctionAdvisor(differentiation_rule, math_function) command returns the differentiation rule for the function.
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Examples
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The variables used by the FunctionAdvisor command to create the calling sequence are local variables. To make the FunctionAdvisor command return results using global variables, pass the actual function call instead of the function name. Compare the following two input and output groups.
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* Partial match of "diff" against topic "differentiation_rule".
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![eq1 := Diff(Zeta(n, f(z), v), z) = Zeta(n+1, f(z), v)*(diff(f(z), z)), Diff(Zeta(n, z, f(v)), v) = -(z*Zeta(n, z+1, f(v))+(PIECEWISE([0, n = 0], [n*Zeta(n-1, z+1, f(v)), otherwise])))*(diff(f(v), v))](/support/helpjp/helpview.aspx?si=7478/file01727/math81.png)
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![has([eq1], a), has([eq1], b), has([eq1], z)](/support/helpjp/helpview.aspx?si=7478/file01727/math85.png)
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* Partial match of "diff" against topic "differentiation_rule".
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![eq2 := Diff(Zeta(a, f(b), z), b) = Zeta(a+1, f(b), z)*(diff(f(b), b)), Diff(Zeta(a, b, f(z)), z) = -(b*Zeta(a, b+1, f(z))+(PIECEWISE([0, a = 0], [a*Zeta(a-1, b+1, f(z)), otherwise])))*(diff(f(z), z))](/support/helpjp/helpview.aspx?si=7478/file01727/math98.png)
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![has([eq2], a), has([eq2], b), has([eq2], z)](/support/helpjp/helpview.aspx?si=7478/file01727/math102.png)
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For functions which accept different numbers of parameters, you can specify for which function call you want the differentiation rule by specifying the function with the appropriate number of arguments. For example, for Zeta, if given with only one argument specified, it represents the Hurwitz Zeta function and its differentiation rule is the following.
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* Partial match of "diff" against topic "differentiation_rule".
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![Diff(Ei(f(a), z), a) = -(diff(f(a), a))*z^f(a)*MeijerG([[], [0, 0]], [[-1, -1, -f(a)], []], z), Diff(Ei(a, f(z)), z) = -(diff(f(z), z))*Ei(a-1, f(z))](/support/helpjp/helpview.aspx?si=7478/file01727/math134.png)
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As another example, consider the exponential integral Ei.
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![Diff(Ei(f(a), z), a) = -(diff(f(a), a))*z^f(a)*MeijerG([[], [0, 0]], [[-1, -1, -f(a)], []], z), Diff(Ei(a, f(z)), z) = -(diff(f(z), z))*Ei(a-1, f(z))](/support/helpjp/helpview.aspx?si=7478/file01727/math153.png)
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