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anonymous

  • 5 years ago

find the derivative of: y= ln[(3x-1)^3 (4x+2)^5)

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  1. amistre64
    • 5 years ago
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    y = ln(u^3 v^5) u = 3x-1 ; v=4x+2 or y = ln(uv) ; u = (3x-1)^3 v=(4x+2)^5

  2. amistre64
    • 5 years ago
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    eiher way uv is on the bottom ans the top works out to be u'v' right?

  3. anonymous
    • 5 years ago
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    so i have to use the power rule?

  4. amistre64
    • 5 years ago
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    for the top yes; Dx(uv)

  5. anonymous
    • 5 years ago
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    thank yos so much. i just get confused when choosing which rule to use. =D

  6. amistre64
    • 5 years ago
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    uv' + u'v the top is: [(3x-1)^3] [20(4x+2)^4] + [9(3x-1)^2][(4x+2)^5] might wanna clean that up some tho...

  7. anonymous
    • 5 years ago
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    im going to work it out using the formula n then im going to check it up with your answer. thank you so much ^.^

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