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swissgirl

  • 3 years ago

What is the fourth derivative of \(f(x)=3xe^x-e^{2x}\) ?

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  1. bahrom7893
    • 3 years ago
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    well f'(x*e^x) = x*e^x + e^x = e^x(x+1). So: f'(x) = 3e^x(x+1) - 2e^(2x) f''(x) = (3xe^x+3e^x - 2e^(2x))' = 3e^x(x+1)+3e^x - 4e^(2x)

  2. bahrom7893
    • 3 years ago
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    f'''(x) = (3xe^x + 3e^x + 3e^x - 4e^(2x))' = (3xe^x + 6e^x - 4e^(2x))' = 3e^x(x+1) + 6e^x - 8e^(2x)

  3. bahrom7893
    • 3 years ago
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    I think I made a mistake in f''(x)

  4. Hero
    • 3 years ago
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    Yes, you did

  5. bahrom7893
    • 3 years ago
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    No i didn't

  6. Hero
    • 3 years ago
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    You made a mistake somewhere

  7. bahrom7893
    • 3 years ago
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    f''''(x) = (3xe^x+3e^x+6e^x-8e^(2x))' = (3xe^x + 9e^x - 8e^(2x))' = 3e^x(x+1)+9e^x - 16e^(2x) <-Final answer

  8. swissgirl
    • 3 years ago
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    Btw u didnt go wrong

  9. swissgirl
    • 3 years ago
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    Thankkksss Bahhhrrrooommmmmmm

  10. bahrom7893
    • 3 years ago
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    Hero.. like i said.. ignore me.

  11. Hero
    • 3 years ago
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    Yeah, that's the correct answer

  12. Hero
    • 3 years ago
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    Why do you want me to ignore you @bahrom7893

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