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d/dx [f(x)]^n

???

THANK U!

Guest Nov 20, 2018
 #1
avatar+14579 
0

Hmmmm     not sure...        n f(x)^(n-1)   ???

ElectricPavlov  Nov 20, 2018
 #2
avatar+92814 
+1

This is known as the "Chain Rule"

 

d/dx   [ f(x) }^n   =

 

n * [ f(x)] ^(n - 1) * [ f ' (x) ]

 

Where f ' x   is the derivative of f(x)

 

 

cool cool cool

CPhill  Nov 20, 2018
 #3
avatar+14579 
+1

Yikes....thanx Chris !

ElectricPavlov  Nov 20, 2018
 #4
avatar+92814 
0

HAHAHA!!!!!....don't feel bad.....I took 4 semesters of Calculus and can barely do a derivative, now   !!!!

 

 Use it... or Lose It.......!!!!

 

cool cool cool

CPhill  Nov 20, 2018
 #5
avatar+14579 
+1

Yah....I had cal 1    cal 2   cal 3    Diffy q   Advanced Applied Math....but it was 35 years ago, with no practice since....

    BUT  I get to re-learn a bit here.....!

ElectricPavlov  Nov 20, 2018
 #6
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0

All that advanced math, buried and decomposing for 35 years. Along comes the internet and a math forum, and you have nothing better to do, so you decide to dig up the partially mummified, desiccated remains and display them on here, in all their ignominious glory. 

 

Are you hoping it rejuvenates and comes back to life?

Guest Nov 20, 2018
 #7
avatar+1225 
0

Hummm… Though I’m quite capable writing a troll post like this… for the record, I didn’t create this post.

There is another troll on here, one with gorilla DNA and dubious genetic enhancement, who usually wields blunt instruments of curse-words and vulgarities, but sometimes he sharpens his teeth and bites. He usually does this when trolling the more intelligent members and guests.  

 

Which ever method he uses, his posts are usually amusing (though not always). This one is a long way from his masterpiece, but it still ranks above average.  Out of respect for the particular target(s) of this post, I’m keeping my laughter below the threshold of LMAO… It’s not easy though!laughsmiley

 

GA

GingerAle  Nov 21, 2018

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