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Math Help - A couple of logarithmic differentiation questions

  1. #1
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    A couple of logarithmic differentiation questions

    I'm not sure how to derive these:

    y = log(2)x^(1/3) (2)=base

    and

    y = √x(e^3x)
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  2. #2
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    Quote Originally Posted by b521 View Post
    I'm not sure how to derive these:

    y = log(2)x^(1/3) (2)=base

    and

    y = √x(e^3x)
    To your first question:

    Rewrite the given equation:

    y = \log_2(x^{\frac13})~\implies~y=\frac13 \cdot \frac1{\ln(2)} \cdot \ln(x)

    You only have to know how \ln(x) is differentiated, all other factors are constant.

    Do the second question similarly.
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  3. #3
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    Only ln(x) is differentiated? So it would be 1/3xln2 ?
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  4. #4
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    Well, with parentheses to make it clearer: 1/(3(ln 2)x). Yes, that is correct.

    In general, the derivative of e^x is e^x while the derivative of a^x with a any positive number, is a^x ln(a). The derivative of ln(x) is \frac{1}{x} while the derivative of log_a(x) with a any positive number is \frac{1}{x ln(a)}.
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  5. #5
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    So for the second equation, would the answer use fg'+gf'?

    6ln(3x)x^(1/2) + 1/(2x^(1/2))e^(3x)
    (6√x)ln(3x) + e^(3x)/(2√x)

    Is this answer correct?
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  6. #6
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    Quote Originally Posted by b521 View Post
    So for the second equation, would the answer use fg'+gf'?

    6ln(3x)x^(1/2) + 1/(2x^(1/2))e^(3x)
    (6√x)ln(3x) + e^(3x)/(2√x)

    Is this answer correct?
    1. I assume that the given function reads:

    y = \sqrt{x} \cdot e^{3x^2}

    2. Re-write so that you have only one power to the base e at the RHS:

    y =  e^{3x^2+\frac12 \ln(x)}

    3. Now differentiate using the chain rule:

    y'= e^{3x^2+\frac12 \ln(x)} \cdot \left(6x+\frac1{2x}\right)

    Simplify!

    4. If you want to use the product rule and the chain rule you'll get:

    y = \sqrt{x} \cdot e^{3x^2}~\implies~y'=e^{3x^2} \cdot \frac1{2\sqrt{x}} + \sqrt{x} \cdot e^{3x^2} \cdot 6x

    Simplify!
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