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04292013  #1 
Join Date: Apr 2013
Posts: 1

First Principle of Derivative and Power Rule
using the first principle of differentiation, find the first derivatives of
1.f(x) = 3/x^2 2. 1/(sqrt x)^3 i don't know how to using limit to solve this. Please help me. 
04302013  #2  
Join Date: Dec 2008
Posts: 249

Quote:
f'(x) = ((derivative of the top)(bottom)  (top)(derivative of the bottom))/(bottom squared) f'(x) = ((0)(x^2)  (3)(2x))/(x^4) = 6x/x^4 = 6/x^3 Using the product rule to solve: 3/x^2 = 3x^2 f'(x) = (derivative of first)(second) + (first)(derivative of second) f'(x) = (0)(x^2) + (3)(2x^3) = 6x^3 = 6/x^3 Using limit method: limit as h goes to 0 of (3/(x+h)^2  3/x^2)/h = ((3x^2  3(x + h)^2)/((x^2)(x + h)^2))/h = (3x^2  3x^2  6xh  3h^2)/((h)(x + h)^2(x^2)) = (3h(2x + h))/((h)(x + h)^2(x^2)) = (3(2x + h))/((x + h)^2(x^2)) Now take the limit as h goes to 0. = (6x)/(x^4) = 6/x^3 2. f(x) = 1/(sqrt x)^3 = (sqrt(x))^3 = (x^(1/2))^3 = x^(3/2) f'(x) = (3/2)(x^(3/2  1) = (3/2)(x^(5/2) = 3/(2(sqrt(x))^5) Last edited by MAS1; 04302013 at 10:48 AM.. Reason: Adding limit method 

05012013  #3 
Join Date: Mar 2013
Posts: 2

sorry
sorry i don't now that you out of luck

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