find the derivative
f(x)=(c^9-x^9)/(c^9+x^9), c is constant

Question

find the derivative 
f(x)=(c^9-x^9)/(c^9+x^9), c is constant

geerky42 · Answer

What do you need help with?

anonymous · Answer

its calculus find the derivation which i know how to do but i have no clue what c is constant means

geerky42 · Answer

c is just a unknown number. treat it as number.

campbell_st · Answer

image it's  512 or 2^9   that's all it means... some arbitrary number

anonymous · Answer

if i just take the derivation of that equation though using the quotient formula it will equal out to 0 which is wrong

freckles · Answer

derivation?

freckles · Answer

derivative?

anonymous · Answer

yes derivative sorry

freckles · Answer

would you know how to find the derivative of say something like:
$$f(x)=\frac{1^9-x^9}{1^9+x^9}=\frac{1-x^9}{1+x^9}$$

anonymous · Answer

yes i just use the quotient rule I just don't understand what c being constant means

anonymous · Answer

or i can use the chain rule to figure it out

freckles · Answer

like c doesn't change 
c doesn't vary 
it isn't a variable 
derivative of c is 0

freckles · Answer

just like the derivative of 1 is 0

geerky42 · Answer

http://www.mathsisfun.com/definitions/constant.html

freckles · Answer

1 is actually called a constant because it constantly remains 1

geerky42 · Answer

$\dfrac{d}{dx}c = 0$

freckles · Answer

sqrt(30) is a constant

freckles · Answer

i'm sure you heard of rules like constant multiple rule and constant rule if you are being asked to differentiate this

freckles · Answer

$$(cf)'=c(f')$$
means if c is some number like 5 or 3 or -2 or -1/2 or sqrt(3)
then you can factor it out and bring down side the derivative of the part that varies

anonymous · Answer

actually no my teacher is a ta thats not so great lol i teach myself everything

freckles · Answer

and that was the constant multiple rule i just mentioned

freckles · Answer

the constant rule is derivative of a constant being 0

anonymous · Answer

so basically i replace c with 0

freckles · Answer

$$f(x)=\frac{c^9-x^9}{c^9+x^9} \ f'(x)=\frac{(c^9-x^9)'(c^9+x^9)-(c^9+x^9)'(c^9-x^9)}{(c^9+x^9)^2}$$

so basically since c is a constant c^9 is a constant
and derivative of c^9 would be 0

anonymous · Answer

and if i do that the derivative will still all equal to 0 which web assign is telling me is wrong because i originally thought that you would take c equal it to 0 but when web assign said it was wrong i just assume I didn't know what c being constant really meant

freckles · Answer

no the derivative of all of that is not equal to 0

freckles · Answer

what is derivative of (c^9-x^9)?

anonymous · Answer

9c^8-9x^8

freckles · Answer

$$\frac{d}{dx}(c^9-x^9)=\frac{d}{dx}c^9 -\frac{d}{dx} x^9$$
what does this equal?

freckles · Answer

not exactly

freckles · Answer

again c is a constant

freckles · Answer

$$\frac{d}{dx}(c^9-x^9)=\frac{d}{dx}c^9 -\frac{d}{dx}x^9=0-9x^8 $$

anonymous · Answer

well if you are talking about it being constant then c would be 0 so it will be -9x^8

freckles · Answer

well c wouldn't be zero but the derivative of c is zero

anonymous · Answer

ok i get it now thanks

freckles · Answer

$$f(x)=\frac{c^9-x^9}{c^9+x^9} \ f'(x)=\frac{(c^9-x^9)'(c^9+x^9)-(c^9+x^9)'(c^9-x^9)}{(c^9+x^9)^2}  \ f'(x)=\frac{(-9x^8)(c^9+x^9)-(c^9+x^9)'(c^9-x^9)}{(c^9+x^9)^2}$$

anonymous · Answer

n

freckles · Answer

so you still need to differentiate (c^9+x^9)

freckles · Answer

with respect to x of course

anonymous · Answer

yes i understand it now i just solved it and it was right thank you

freckles · Answer

ok cool

freckles · Answer

i said c isn't zero
c is a constant and it could be 0
but they want the derivative formula for any constant c

freckles · Answer

and ok i know you get 
so i will shut up now

anonymous · Answer

lol its k thank you again