## A maths joke I don't get

For the discussion of math. Duh.

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jamespellis
Posts: 3
Joined: Wed May 25, 2011 4:14 pm UTC

### A maths joke I don't get

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

Gwydion
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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:
d/dy, as ce^x is not a varying function of y and therefore the partial derivative is 0. The "joke" coming from the fact that d(e^x)/dx is e^x, so e^x is "invincible" to differential operators.

drachefly
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### 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?

jamespellis
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Joined: Wed May 25, 2011 4:14 pm UTC

### 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).

notzeb
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### 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 d2/dc2, and then it would be 0.

Or you could differentiate it once wrt c, and then it would be e^x.
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BedderDanu
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Joined: Tue Jan 14, 2014 6:18 am UTC

### Re: A maths joke I don't get

Assuming I'm remembering my DiffEQ class correctly:
_Dx_
. c-x
Spoiler:
Because it makes cex discontinuous.

...I'll see myself out.

notzeb
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### 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.)
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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...

Code: Select all

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