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[-] PartySlices@lemmy.world 21 points 6 days ago

Reminds me of a fun joke!

A constant and e^x were walking down the street. They see a differentiator coming, and instantly the constant gets scared shitless. The differentiator comes up, and as expected, POOF, the constant is gone. e^x starts laughing and says "nice try, can't touch me". But the differentiator tips his hat and responds "is that right?", then throws off his coat and shows that he is d/dy.

[-] ChaoticNeutralCzech@lemmy.ml 3 points 6 days ago* (last edited 6 days ago)

Nobody exp(x) the y-derivation!

[-] ChaoticNeutralCzech@feddit.org 3 points 6 days ago

The result is ๐‘ฆโ€Š=โ€Šโ…Ÿโ‚“, right?

[-] Papergeist@lemmy.world 24 points 6 days ago

No e^x doesn't have a 'y' and so it also acts as a constant.

[-] ChaoticNeutralCzech@feddit.org 9 points 6 days ago* (last edited 6 days ago)

Oh, I was thinking of it as ๐‘ฆโ€‰= ๐‘’^๐‘ฅ^ or ๐‘ฅโ€‰= ln ๐‘ฆ, whose derivative in respect to ๐‘ฆ is ๐‘ฅโ€‰= 1/๐‘ฆ (for ๐‘ฆโ€‰> 0) or ๐‘ฆโ€‰= 1/๐‘ฅโ€‰(for ๐‘ฅ > 0). Your interpretation is that the ๐‘ฆ-axis is non-existent or named differently, which is why I'd prefer the joke to say d/d๐‘ก for less ambiguity, as @anothercatgirl suggested.

yayaya, or in other cases like multiple independent variables, I'm not sure because it's been 6 years since I took calculus

[-] MOCVD@mander.xyz 2 points 5 days ago

The general form would be implicit differentiation! d/dx dx/dy e^x = e^x dx/dy

this post was submitted on 08 Nov 2025
615 points (99.0% liked)

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