Question

use the definition of derivative to find f'(x) for f(x)=(3-5x)^1/2

Answer 1

limit definition is f'(x) = lim as h -> 0 [f(x+h) - f(x)]/h

Answer 2

plug that in:
lim as h -> 0 {(3 - 5(x+h))^(1/2) - (3-5x)^(1/2)}/h

Answer 3

now to get rid of square roots, use the identity a^2 - b^2 = (a-b)(a+b)
here you have a - b, but are missing a + b. So multiply top and bottom by:
(3 - 5(x+h))^(1/2) + (3-5x)^(1/2)

Answer 4

lim as h -> 0 [ (3 - 5(x+h))^(1/2) - (3-5x)^(1/2) * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ]/[ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ]

Answer 5

that is equal to the:
lim as h -> 0 [ (3 - 5(x+h)) - (3-5x) ] / [ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ]

Answer 6

simplify numerator:
lim as h -> 0 of [ (3 - 5(x+h)) - (3-5x) ] / [ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ] =
= lim as h -> 0 of [ 3 - 5x-5h - 3+5x) ] / [ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ] =
= lim as h -> 0 of [-5h]/[ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ]

Answer 7

Cancel out "h"s
lim as h -> 0 of [-5h]/[ h * (3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ] =
lim as h -> 0 of [-5]/[(3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ]

Answer 8

Plug in h = 0:
lim as h -> 0 of [-5]/[(3 - 5(x+h))^(1/2) + (3-5x)^(1/2) ] = [-5]/[(3 - 5(x)^(1/2) + (3-5x)^(1/2) ]

Answer 9

As 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 =)

Answer 10

thanx for ur reply