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Funniest math joke EVAR

beer

Lifer
A constant function and e^x are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. So e^x follows him and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!" "Ah," says e^x, "he won't bother ME, I'm e to the x!" and he walks on. Of course he meets the differential operator after a short distance.

e^x: "Hi, I'm e^x"

diff.op.: "Hi, I'm d/dy"
 
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?
 
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng

He is differentiating d/dy, not d/dx.

It's not all that funny, IMO, but it's pretty good.
 
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

ooooh hehe i get it 😀
 
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

d/dy(e^x) = e^x*x'
 
Sounds like some people need to retake calc 1. d/dy is differentiation with respect to y. Since e^x is a function of x, d(e^x)/dy is 0.
 
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

iirc d(e^x)/dy = e^x, so e^x wasn't worried. but i still don't get the convo at the end either.

EDIT: i just read the other posts. now i get it. sheesh it's been a long time since calc.
 
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

oops 😀
 
Originally posted by: XZeroII
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

d/dy(e^x) = e^x*x'

No
You are differentiating with respect to Y
therefore, you treat e^x as a constant
d/dy of any constant is zero
 
you know that reminds me of.

For those of you who watch Cartoon Network. Dexter Tells this Elabert math joke that makes no blood sense but you can't help but laugh because it's so stupid.
 
Originally posted by: Elemental007
Originally posted by: XZeroII
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

d/dy(e^x) = e^x*x'

No
You are differentiating with respect to Y
therefore, you treat e^x as a constant
d/dy of any constant is zero

no, you must implicitly differentiate.
 
Originally posted by: Elemental007
Originally posted by: XZeroII
Originally posted by: Haircut
Originally posted by: gopunk
i don't get the conversation they have at the end... am i supposed to sound it out or sometihng
No, think of the fact that e^x was cocky about how getting differentiated couldn't do anything to him

Now, what is d/dy (e^x)?

d/dy(e^x) = e^x*x'

No
You are differentiating with respect to Y
therefore, you treat e^x as a constant
d/dy of any constant is zero

actually i thought that. but his way is also correct, because x could be a function of y, and it yields the correct answer (0) if it is constant.
 
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