# Find the second derivate .(tricky question).

by Penultimate
Tags: derivate, tricky
 P: 26 There Fx a twice derivable function. Find Y'' if $$Y=f(\frac{2}{x})$$
 PF Gold P: 7,120 You need to show some work first. Do you know the chain rule?
 P: 26 I have problems with english ezpresions about math and i am not sure what you refer with chain rule.
 PF Gold P: 7,120 Find the second derivate .(tricky question). The "chain rule" can be summarized as $$\frac{{df}}{{dx}} = \frac{{df}}{{du}}\frac{{du}}{{dx}}$$. Make the substitution $$u = \frac{2}{x}$$ and you can take derivatives of f(2/x). Remember, f(2/x) is any arbitrary function with 2/x as the argument so when you find your result, don't be surprised that you have arbitrary derivatives, that is terms including f'(2/x) and f''(2/x).
 P: 3,016 Penultimate: You have at least 5 or 6 of these threads. No attempts at anything in any of them. We do not do your work for you, regardless of "how much time you have or do not have ." Don't be lazy and show some effort.
 P: 26 Saladsamurai i know i and wish could but yesterday i got these question and tomomorow i have got the analysis exam. I dont expect you to do my homework , but i still hope you would solve the problem i have submited. I dont know what else to say , whatever you will do with my threads i thank you again for the patience you are showing. Good night everybody have a nice time.
P: 3,016
 Quote by Penultimate Saladsamurai i know i and wish could but yesterday i got these question and tomomorow i have got the analysis exam. I dont expect you to do my homework , but i still hope you would solve the problem i have submited. I dont know what else to say , whatever you will do with my threads i thank you again for the patience you are showing. Good night everybody have a nice time.
You are again missing the point. We do not solve HW at PF. If you read the forum rules, you would know that.

If you show some effort, we will guide you through the solutions.

Good luck on your exam. And when you are ready to learn how to solve these and not just have them solved for you, come on back.

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