Does anyone "get" this joke?
"Let f(x) = c and g(x) = e^x. The product function fg is afraid of which differential operator?"
It was a joke made by my lecturer in some online notes and nobody I know can figure it out
A maths joke I don't get
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Re: A maths joke I don't get
It's a really lame joke, but I think the answer is probably
Spoiler:
Re: A maths joke I don't get
I'm not sure why they split it into two functions, then.
d/d(anything with no partial in respect to c and x) would leave them 0.
d/d(e^x) would result in c, a.k.a, f(x), and vice versa, but I don't see comedic potential.
f(x)g(x) = ce^x, which looks like it ought to be able to pun off of sex somehow, but not so much with 'fear'.
Are you sure this has been reproduced faithfully?
d/d(anything with no partial in respect to c and x) would leave them 0.
d/d(e^x) would result in c, a.k.a, f(x), and vice versa, but I don't see comedic potential.
f(x)g(x) = ce^x, which looks like it ought to be able to pun off of sex somehow, but not so much with 'fear'.
Are you sure this has been reproduced faithfully?

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Re: A maths joke I don't get
My best thought was a "second hand derivative" due to cex being a second hand type shop in the UK (where I live), but it doesn't explain being afraid.
I also considered sex jokes and came up short (I'm better at them clearly).
Partially differentiating by y may be the answer, but I'll be disappointed if it is.
It has been reproduced faithfully, I actually copied and pasted it from the 2013 exam for this course (I kid you not) where the answer was awarded 1 mark (no mark scheme given unfortunately).
I also considered sex jokes and came up short (I'm better at them clearly).
Partially differentiating by y may be the answer, but I'll be disappointed if it is.
It has been reproduced faithfully, I actually copied and pasted it from the 2013 exam for this course (I kid you not) where the answer was awarded 1 mark (no mark scheme given unfortunately).
Re: A maths joke I don't get
(d/dx  1) kills it, and doesn't rely on silly conventions like the d/dy answer (not everybody calls the output of a function "y", and for all we know x might be a function of y).
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 jestingrabbit
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Re: A maths joke I don't get
The fact that cex is a second hand store is interesting. You could differentiate wrt to c twice, which would be d^{2}/dc^{2}, and then it would be 0.
Or you could differentiate it once wrt c, and then it would be e^x.
Or you could differentiate it once wrt c, and then it would be e^x.
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Re: A maths joke I don't get
Assuming I'm remembering my DiffEQ class correctly:
_D_{x}_
. cx
_D_{x}_
. cx
Spoiler:
Re: A maths joke I don't get
There's also the trivial answer: 0. It's the differential operator that takes any function as input, and returns the (constant) 0 function as output.
Alternative answer: maybe by fg, your lecturer meant f∘g. Since (f∘g)(x) = c for all x, d/dx kills it. (So does 0, of course.)
Alternative answer: maybe by fg, your lecturer meant f∘g. Since (f∘g)(x) = c for all x, d/dx kills it. (So does 0, of course.)
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 phlip
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Re: A maths joke I don't get
I'm not entirely sure what the joke in the OP is supposed to mean on its own, but it would make at least some amount of sense as a followup to this classic joke. Still not entirely sure what it would mean, but there would at least be some context to be had...
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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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