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ammu123

  • one year ago

find the derivative of y with respect to x, t or theta. (a) e^7 - 10x Answer: -10e^7-10x (b) 8xe^x - 8e^x answer: 8xe^x (c) y= (x^2 -2x+4)e^x answer: (x^2+2)e^x (d) y= sin e^theta^4 answer: (-4theta^3 e^-theta^4) cos e^-theta^4 please show the steps. thank you

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  1. jim_thompson5910
    • one year ago
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    The first one is \[\Large e^{7-10x}\] right?

  2. ammu123
    • one year ago
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    yes

  3. jim_thompson5910
    • one year ago
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    The derivative of that isn't -10e^7-10x, so there has to be a typo somewhere

  4. ammu123
    • one year ago
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    im not sure i think we have to use the u substitution method to find the answer.

  5. jim_thompson5910
    • one year ago
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    oh wait, is the answer is \[\Large -10e^{7-10x}\] and not \[\Large -10e^{7}-10x\]

  6. ammu123
    • one year ago
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    yea its the first one

  7. jim_thompson5910
    • one year ago
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    this is why parenthesis are useful if 7-10x is in the exponent, then say e^(7-10x)

  8. jim_thompson5910
    • one year ago
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    ok that makes more sense now

  9. ammu123
    • one year ago
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    sorry about that

  10. jim_thompson5910
    • one year ago
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    let u = 7 - 10x, which means du/dx = -10 this means \[\Large e^{7-10x}\] turns into \[\Large e^{u}\] then derive to get \[\Large e^{u}*\frac{du}{dx}\] \[\Large e^{u}(-10)\] \[\Large -10e^{u}\] \[\Large -10e^{7-10x}\]

  11. jim_thompson5910
    • one year ago
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    I'm using the chain rule, which is If h(x) = f(g(x)), then h ' (x) = f ' (g(x)) * g ' (x)

  12. ammu123
    • one year ago
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    ok i just need help with the other 3 as well

  13. jim_thompson5910
    • one year ago
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    whats the derivative of 8xe^x

  14. ammu123
    • one year ago
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    im not sure

  15. jim_thompson5910
    • one year ago
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    use the product rule

  16. jim_thompson5910
    • one year ago
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    tell me what you get

  17. ammu123
    • one year ago
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    is it 8xe^x

  18. jim_thompson5910
    • one year ago
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    no, but that's part of it though

  19. jim_thompson5910
    • one year ago
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    Product Rule: if h(x) = f(x)*g(x), then h ' (x) = f ' (x) * g(x) + f(x) * g ' (x)

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