differentiate y=ln(ln x)?

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  • MyRank
    Lv 6
    2 weeks ago

    y = log (logx)

    dy/dx = 1/ logx d/dx (logx)= 1/ logx x 1/x= 1/ xlogx.

  • 2 weeks ago

    y = ln(lnx)

    dy/dx = d/dx(lnx)/lnx

    dy/dx = (1/x)/lnx

    dy/dx = 1/(xlnx)

  • 2 weeks ago

    d/dx lnu = u'/u

    u' = 1/x

    (1/x)/lnx

    1/xlnx

  • hfshaw
    Lv 7
    2 weeks ago

    You have:

    y = ln(ln x)

    There are many ways to find dy/dx (e.g., chain rule, implicit differentiation...). Let's use implicit differentiation.

    Exponentiate both sides:

    exp(y) = ln(x)

    Differentiate both sides with respect to x:

    exp(y)*dy/dx = 1/x

    solve for dy/dx:

    dy/dx = exp(-y)/x

    Back substitute for exp(y) [if exp(y) = ln(x), exp(-y) = 1/ln(x)]:

    dy/dx = 1/(x*ln(x))

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  • 2 weeks ago

    Find the derivative of the following via implicit differentiation:

    d/dx(y) = d/dx(ln(ln(x)))

    Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):

    d/dx(x) y'(x) = d/dx(ln(ln(x)))

    The derivative of x is 1:

    1 y'(x) = d/dx(ln(ln(x)))

    INTERMEDIATE STEPS:

    Possible derivation:

    d/dx(x)

    Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1.

    d/dx(x) = d/dx(x^1) = x^0:

    Answer: |

     | = 1 |

    Using the chain rule, d/dx(ln(ln(x))) = ( dln(u))/( du) ( du)/( dx), where u = ln(x) and d/( du)(ln(u)) = 1/u:

    y'(x) = (d/dx(ln(x)))/ln(x)

    Using the chain rule, d/dx(ln(x)) = ( dln(u))/( du) ( du)/( dx), where u = x and d/( du)(ln(u)) = 1/u:

    y'(x) = ((d/dx(x))/x)/ln(x)

    Simplify the expression:

    y'(x) = (d/dx(x))/(x ln(x))

    The derivative of x is 1:

    Answer: y'(x) = 1 1/(x ln(x))

  • 2 weeks ago

    y'(x) = 1/(x log(x))

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