Here's the question you clicked on:
mathcalculus
DERIVATIVES: (attached below)
Now that's one that confuses me, too, lol x_x Not even sure what you're supposed to do with it. Curious to see what other people say.
Oh, wait, I think it just wants the function?
we can all use the help.
Can We Use L-hospitals Rule ?
Don't even know how to do that :P But the only thing I would be able to do is recognize that the function is x^(-3.25) and go from there. But that;d be cheating I think. And nah, l-hopitals rule is above his class.
i don't know. it's a derivative question.
L'hopitals rule is a calc II thing, so not much to worry about for yourself, lol.
If I were you i would use hopital phenomenal .
yeah, no no hospital rule.
differentiate both numerator n denominator with respect to h u got ur answer n put h =0
I was thinking the same as Psyman mentioned that the original function is y=x^(-3.25) dy/dx=-3.25x^(-2.25) i guess in order to evaluate the limit Binomial expansion is required
simple l hopital rule wud do in this case
I was thinking the expansion wasn't what the question really wanted, though.
He is Just Newbie in the calculus I think He dont know L'Hôpital's rule !
Yeah, he hasn't seen l'hopitals rule if he's just now doing this.
SIGN: people who know this only.
\[\frac{ dx^p }{ dx }=\lim_{h \rightarrow 0}\frac{ (x+h)^p-x^p }{ h }=px^{p-1}\] And in this case p=-3.125
Then the solution:\[-3.25x^{-4.25}\]
this is as mentioned above the derivative of the function \[\Large y=x^{-3.25}\] using the power rule of derivative which states if \[\Large y=x^n\] \[\Large \frac{dy}{dx}=nx^{n-1}\] here n=3.25 just appply the power rule formula \[\Large \frac{dy}{dx}=-3.25x^{(-3.25-1)}\] \[\Huge \frac{dy}{dx}=-3.25x^{-4.25}\]
Yeah, more of just a knowledge of what the power rule for a derivative is and where within that difference quotient the original function actually is. In the end, it's the start of getting used to nx^(n-1) for derivatives :P