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hpfan101

  • one year ago

Differentiate: \[\frac{ x^2e^x }{ x^2+e^x }\]

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  1. Empty
    • one year ago
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    Are you familiar with the quotient rule?

  2. UnkleRhaukus
    • one year ago
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    \[\large\left(\frac uv\right)' = \frac{u'v-uv'}{v^2}\]

  3. UnkleRhaukus
    • one year ago
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    @hpfan101, can you identify \(u\) and \(v\)? can you find the derivatives \(u'\) and \(v'\)?

  4. hpfan101
    • one year ago
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    I'm somewhat familiar with the quotient rule.

  5. hpfan101
    • one year ago
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    Would it be: \[\frac{ (x^2e^x)'(x^2+e^x)-(x^2e^x)(x^2+e^x)' }{ (x^2+e^x)^2 }\]

  6. UnkleRhaukus
    • one year ago
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    yeah, that's right, now simplify those derivatives

  7. hpfan101
    • one year ago
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    Okay, what I'm not sure about is the first set we would find the derivative of. Would the derivative of \[(x^2e^x)'\] be \[2xe^x\]?

  8. UnkleRhaukus
    • one year ago
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    use the product rule \[(fg)' = f'g+fg'\]

  9. hpfan101
    • one year ago
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    Oh okay. So it would be: \[(x^2)'(e^x)+(x^2)(e^x)'= 2x(e^x)+x^2(e^x) =2xe^x+x^2e^x\]

  10. UnkleRhaukus
    • one year ago
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    that's right and if you factor it \(= (2x+x^2)e^x\)

  11. hpfan101
    • one year ago
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    Oh okay! Thank you! I think I can figure out the rest.

  12. UnkleRhaukus
    • one year ago
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    to compute\[(x^2+e^x)'\] use the sum rule \[(a+b)'=a'+b'\]

  13. hpfan101
    • one year ago
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    Okay, will do! Thanks!

  14. UnkleRhaukus
    • one year ago
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    NB: expanding the square in the denominator will not simplify your result (in this case)

  15. hpfan101
    • one year ago
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    Ah, okay. I see. It'd be easier to leave the denominator how it is.

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