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anonymous

  • 5 years ago

estimate (.99e^.02)^8 with linear approximation.

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  1. anonymous
    • 5 years ago
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    I'm getting lost trying to apply the formula fx (a,b) delta x + fy (a,b) delta y

  2. anonymous
    • 5 years ago
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    I know that e^0 = 1 but I get confused when trying to apply this to the f(x) function to take the derivative of it. so it would be F(x) = (.99e^x)^8 take the deriv of this?

  3. watchmath
    • 5 years ago
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    Consider the function \((xe^y)^8\) and you want to estimate that near (1,0)

  4. anonymous
    • 5 years ago
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    ohhh, so would I take partial derivs? or am I think of the wrong thing. Because I believe using that formula i menitoned before it delta y =.01 delta x=.02

  5. watchmath
    • 5 years ago
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    Yes, you take the partial derivative and delta y = -.01

  6. watchmath
    • 5 years ago
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    To make it nicer you probably want to write the function as \(x^8e^{8y}\)

  7. anonymous
    • 5 years ago
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    i figured it out! did the easy derivative wrong haha

  8. amistre64
    • 5 years ago
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    :) it happens

  9. anonymous
    • 5 years ago
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    \[(1+((8(1)^7)*(-.01)) + (0+8e^0)*(.02))\] which =1.08 which is a much better percentage of error.

  10. anonymous
    • 5 years ago
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    quaint pre-calculator problem from math teachers youth

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