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anonymous

  • 5 years ago

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

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  1. bahrom7893
    • 5 years ago
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    limit definition is f'(x) = lim as h -> 0 [f(x+h) - f(x)]/h

  2. bahrom7893
    • 5 years ago
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    plug that in: lim as h -> 0 {(3 - 5(x+h))^(1/2) - (3-5x)^(1/2)}/h

  3. bahrom7893
    • 5 years ago
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    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)

  4. bahrom7893
    • 5 years ago
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    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) ]

  5. bahrom7893
    • 5 years ago
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    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) ]

  6. bahrom7893
    • 5 years ago
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    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) ]

  7. bahrom7893
    • 5 years ago
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    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) ]

  8. bahrom7893
    • 5 years ago
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    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) ]

  9. bahrom7893
    • 5 years ago
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    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 =)

  10. anonymous
    • 5 years ago
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    thanx for ur reply

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