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anonymous

  • one year ago

find the derivative of the function h(x)=sqrt(2-x)

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  1. myininaya
    • one year ago
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    use chain rule

  2. myininaya
    • one year ago
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    by the sqrt( ) can be written as ( )^(1/2)

  3. anonymous
    • one year ago
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    we are just in the first section and haven't learned the chain rule, and are required to use a longer method

  4. myininaya
    • one year ago
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    ok the definition then

  5. anonymous
    • one year ago
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    right

  6. myininaya
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =f'(x)\]

  7. myininaya
    • one year ago
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    where we have f(x)=sqrt(2-x) and so f(x+h)=sqrt(2-(x+h)) <--here just replaced my x's with (x+h)'s

  8. myininaya
    • one year ago
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    now plug in

  9. myininaya
    • one year ago
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    sometimes a good method when you are trying to evaluate a limit algebraically and you have a difference/sum of radicals rationalizing might help

  10. myininaya
    • one year ago
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    to rationalize recall conjugates

  11. myininaya
    • one year ago
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    for example the conjugate of a-b is a+b and the conjugate of a+b is a-b

  12. myininaya
    • one year ago
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    do you want to try to see what happens if you do the following: \[\lim_{h \rightarrow 0} \frac{\sqrt{2-(x+h)}-\sqrt{2-x}}{h} \cdot \frac{\sqrt{2-(x+h)}+\sqrt{2-x}}{\sqrt{2-(x+h)}+\sqrt{2-x}}\]

  13. anonymous
    • one year ago
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    right, I got it down to -1/sqrt2-x-h +sqrt2-x

  14. myininaya
    • one year ago
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    one sec let me check

  15. myininaya
    • one year ago
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    that looks pretty sweet

  16. myininaya
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{-1}{\sqrt{2-x-h}+\sqrt{2-x}}\]

  17. myininaya
    • one year ago
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    your job is to just replace h with 0

  18. myininaya
    • one year ago
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    any you could do a little magic afterwards to make the answer look a bit more tidy

  19. anonymous
    • one year ago
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    so -1/2sqrt2-x?

  20. anonymous
    • one year ago
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    thank you!

  21. myininaya
    • one year ago
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    \[\frac{-1}{2 \sqrt{2-x}} \text{ is \right!} \]

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