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pottersheep
Please help! "challenge" calc question. We just learned power, product, chain rules...(I have to draw it out)
|dw:1360808280236:dw|
I must evaluate that
I don't know how :(
use the L'hopital hint : differentiate for the numerator and the denominator
We haven't learned questient rule if that is what you are saying?
it just the derivative of a function... u can use the basic formula : y = x^n ----> y ' = n*x^(n-1) y = a (a is constant) ---> y ' = 0
I tried using product rule and I still got zeros in my denominator though
we neednt use the product or quotien here, just deriv each terms
derivative of x^1000 - 1 = ... ? derivative of x - 1 = .... ?
ohhh because what I did was I wrote it as (X^1000-1)(x-1)^-1
derivative of x^1000 - 1 = 1000x^999 derivative of x - 1 = 1?
yes, that's right... now, just put x=1, u will get its answer
no quotient rule? :o
|dw:1360809016288:dw| NO QUOTIENT RULE :)
That's the right answer! Thank you, how come we dont have to use the quotient rule though if i have a fraction?
and there is division
sorry, icant explaind more... my english so bad :)
You're good at english I didnt even notice :) thank you for your time
I'll try to explain why the quotient rule isn't used. What @RadEn described was L'Hopital's rule, which says that the limit of some rational function f(x)/g(x) (a ratio of f to g), where both f and g are differentiable and g(x) is non-zero, is equal to the limit of the ratio of the derivatives of f and g. \[\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}\] The quotient rule itself wasn't used because you do not find the derivative for f(x)/g(x), but rather f(x) and g(x) separately.