Derivative of exponential integral

[ashiklom]

General form:

$$ \frac{d}{dx} E\nu(x) = -E{\nu - 1}(x) $$

Note that:

$$ E_0(x) = \frac{e^{-x}}{x} $$

Therefore, derivative of SpecialFunctions.expint(x) ($E_1(x)$) is:

$$ \frac{d}{dx} E_1(x) = -E_0(x) = -\frac{e^{-x}}{x} $$

Source: Wikipedia

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Comparison is base (fee3857) 97.86% compared to head (fed2f7d) 97.88%.

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[devmotion]

Looks mainly good (it also matches the rules we added to SpecialFunctions, e.g., https://github.com/JuliaMath/SpecialFunctions.jl/blob/ae35d10713a470b852e53e0c79f20252b5572fa9/ext/SpecialFunctionsChainRulesCoreExt.jl#L183). I am pretty sure though that it requires us to update the SpecialFunctions compat entry since IIRC these functions require a relatively recent release of SpecialFunctions. Can you check this?

[ashiklom]

Good catch! Looks like this starts to work (or at least, all the tests pass) starting in SpecialFunctions v1.1. Bumped compatibility accordingly.