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anonymous

  • 5 years ago

f''(x) for f(x) =e ^-x^2

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  1. anonymous
    • 5 years ago
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    f''(x) for \[f(x)=e ^{-x ^{2}}\]

  2. anonymous
    • 5 years ago
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    Chain Rule?

  3. anonymous
    • 5 years ago
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    \[f'(x)=-2xe ^{-x^2}\] \[f''(x)=-2x(-2xe ^{-x^2})+e ^{-x^2}(-2)\] just simplify!

  4. anonymous
    • 5 years ago
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    where did you get the -2?

  5. anonymous
    • 5 years ago
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    by product rule?

  6. anonymous
    • 5 years ago
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    could you explain i understand how to find derivative just not really with e

  7. anonymous
    • 5 years ago
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    you know that the derivative of e^x is e^x right?

  8. anonymous
    • 5 years ago
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    i forgot that

  9. anonymous
    • 5 years ago
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    the derivative of e^u ,where u = f(x), is e^u(f'(x))

  10. anonymous
    • 5 years ago
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    so like for y=(x^2+9)^4 it would be 4(x^2+9)^3

  11. anonymous
    • 5 years ago
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    derivative of exponential function is the same function multiplied by the derivative if the power.. as dichalao said: \[e ^{f(x)}=f'(x)e ^{f(x)}\] then I used the product rule which states that: \[(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)\]

  12. anonymous
    • 5 years ago
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    lim of (x^3-13x+12)/(x^3-14x+15) as x approaches 3 without using L hopital's rule. :( hope you can help me guys.

  13. anonymous
    • 5 years ago
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    well, this is an easy one.. since, as you see, x=3 is a zero of both The numerator and the denominator, (x-3) is a factor of both of them .. so just divide both of them by (x-3), you will get: \[\lim_{x \rightarrow 3} {(x-3)(x^2+3x-4) \over (x-3)(x^2+3x-5)}\] (x-3)/(x-3) is 1, you just directly substitute in the new expression to get: \[\lim_{x \rightarrow 3}{x^2+3x-4 \over x^2+3x-5}={9+9-4 \over 9+9-5}=14/13\]

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