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Can you find the first derivative of f(x)

for x=3.5, it is 14 and for x=20 is 2.36

That's the slope of tangents at that points. (If you did find f'(x) correctly :))

the answer says -24 for 3.5 tho :S

Tho? :? \[f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24 \]

though*.. I dont get it.. what did you do?

@ sabika13 did you get it? any question here?

use the quotient rule here

d(u/v)dx=(v du/dx - u dv/dx)/v^2

can follow from here?

for y= 2x/(x-3),
let u =2x and v=(x-3)
now do the quotient rule

@ sabika13 are you following now here?

ivenever heard of the quotient rule.. or imforgetting..

what does d and v stand for :S

oh i see, are you in algebra class now?

advanced functions

yes, do you know limits?

limits of a function?

ok advance functions, im sure you already done limits of a function

im not sure if idid.. or what that is

\[\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }\]

thanks

YW, good luck now