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anonymous
 5 years ago
use the definition of derivative to find f'(x) for f(x)=(35x)^1/2
anonymous
 5 years ago
use the definition of derivative to find f'(x) for f(x)=(35x)^1/2

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0limit definition is f'(x) = lim as h > 0 [f(x+h)  f(x)]/h

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0plug that in: lim as h > 0 {(3  5(x+h))^(1/2)  (35x)^(1/2)}/h

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0now to get rid of square roots, use the identity a^2  b^2 = (ab)(a+b) here you have a  b, but are missing a + b. So multiply top and bottom by: (3  5(x+h))^(1/2) + (35x)^(1/2)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0lim as h > 0 [ (3  5(x+h))^(1/2)  (35x)^(1/2) * (3  5(x+h))^(1/2) + (35x)^(1/2) ]/[ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0that is equal to the: lim as h > 0 [ (3  5(x+h))  (35x) ] / [ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0simplify numerator: lim as h > 0 of [ (3  5(x+h))  (35x) ] / [ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ] = = lim as h > 0 of [ 3  5x5h  3+5x) ] / [ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ] = = lim as h > 0 of [5h]/[ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Cancel out "h"s lim as h > 0 of [5h]/[ h * (3  5(x+h))^(1/2) + (35x)^(1/2) ] = lim as h > 0 of [5]/[(3  5(x+h))^(1/2) + (35x)^(1/2) ]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Plug in h = 0: lim as h > 0 of [5]/[(3  5(x+h))^(1/2) + (35x)^(1/2) ] = [5]/[(3  5(x)^(1/2) + (35x)^(1/2) ]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0As long as I didn't make any arithmetic/simplification errors, that is the answer, you can check it by taking the derivative directly. Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)
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