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Math Help - Fairly basic natural logs question

  1. #1
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    Fairly basic natural logs question

    Returning to differentiation after a fairly long break and presented with three questions. I'm sure there's a simpler way to do them, but I can't figure it out?

    1. y(x) = ln[(f(x)+g(x))^2] => y(x) = ln[(u)^2]
    y'(x) = 1*[2(f'(x)+g'(x))] / (f(x)+g(x))^2


    2. y(x) = ln(x^a + zx)/(e^x*e^z)
    y(x) = ln (u)/(v)
    y'(x) = (V) / (U) * (V)(U') / (U)(V')
    y'(x) = (V)(U)(V') / (U)(V)(U')

    If that makes sense?

    3. y(x) = f(g(x).u(z))
    y'(x) = (g(x).u(z)).(g'(x))

    Sorry if this seems a bit retarded, I'm finding the textbook a bit lacklustre in parts (Chiang / Wainwright) and statistics is much more my forte!
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  2. #2
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    Re: Fairly basic natural logs question

    1. y = \ln{u^2}

    y' = \frac{2u}{u'}

    2. y = \frac{\ln{u}}{v}

    y' = \frac{\frac{uv}{u'} - v' \cdot \ln{u}}{v^2}

    3. y = f(u \cdot v)

    y' = f'(u \cdot v) \cdot (uv' + vu')
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    Re: Fairly basic natural logs question

    Quote Originally Posted by skeeter View Post
    1. y = \ln{u^2}

    y' = \frac{2u}{u'}
    I thought with natural logs, the derivative of e.g. y= \ln{x^5} =>  y' = 5x^4 / x^5? I.e. it's the derivative over the original?

    Re: #3, oops. it's the chain rule applying to the product rule, so I see why the answer is that.

    Cheers!
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    Re: Fairly basic natural logs question

    Quote Originally Posted by kedaha View Post
    I thought with natural logs, the derivative of e.g. y= \ln{x^5} =>  y' = 5x^4 / x^5? I.e. it's the derivative over the original?

    Re: #3, oops. it's the chain rule applying to the product rule, so I see why the answer is that.

    Cheers!
    y' = \frac{2u'}{u} ... sorry, transposed the prime
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