Calculus - Derivatives (1 Viewer)

[SpiralFlex]

To clear up any confusing the question is,

Evaluate if



Using the quotient rule,









You owe me an Easter egg!

Get me this one:

 

[shady145]

lol if you're differentiating with respect to x, then the answer is 0 ><

 

[hscishard]

^ ??
That egg looks pretty solid, lol

 

[SpiralFlex]

^ ??
That egg looks pretty solid, lol

My egg.

 

[Jokerers]

Hm.... the function is in terms of x... (i know this by looking at how this is expressed, ie f(x)..) So "r" isn't a variable... so if i differentiate with respect to x, ie f'(x), i will get 0..

OR!!!

this is a chain rule... so u differentiate normally and "blance" the "dr" by mutiplying the final result by "dr/dx" (horrible way of explaining it, i know...). If this is the case;

 

[geni3]

massive noobs here

 

[geni3]

lol if you're differentiating with respect to x, then the answer is 0 ><

ohh wait... you do civil , your stupidity is excused.

how can you assume that r doesnt depend on x noobcake

 

[shady145]

ohh wait... you do civil , your stupidity is excused.

how can you assume that r doesnt depend on x noobcake

wasnt specified so i can assume such things bby gurl

 

[jamesfirst]

Chain rule + Quotient rule.

 

[RANK 1]

im confused, theres no x in (r-6)^3-1/(r-6)^3+1

 

[SpiralFlex]

im confused, theres no x in (r-6)^3-1/(r-6)^3+1

The question was typed incorrectly by the original poster. It has been fixed now.

 

[RANK 1]

and wolfram alpha says its zero

 

[ohexploitable]

and wolfram alpha says its zero

Wolfram Alpha is assuming 'r' is a constant.

 

[ohexploitable]

Wolfram Alpha is assuming 'r' is a constant.

This accounts for both possibilities,

If 'r' is a constant, dr/dx=0 then f'(x)=0 as required. If 'r' is dependent on 'x' then the cake is a lie.