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Math Help - Derivatives of Logarithmic Functions

  1. #1
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    Derivatives of Logarithmic Functions

    Find the derivative:
    1) y=x^(x^3)
    2) y=(sin x)^(x^2)

    Find dy/dx implicitly:
    1) 4xy+ln(x^2*y)=7

    Thanks a lot!
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  2. #2
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    Quote Originally Posted by cupcakelova87 View Post
    Find the derivative:
    1) y=x^(x^3)
    2) y=(sin x)^(x^2)
    I suppose you want to find these using logarithmic differentiation, here goes:

    1) y=x^(x^3)
    => ln(y) = ln[x^(x^3)] ..................log both sides
    => ln(y) = (x^3)ln(x) ....................use a law of logs to bring the power down, log(x^n) = nlog(x)
    => y'/y = 3(x^2)ln(x) + x^2 ..........differentiate both sides with respect to x, use product rule for the right hand side
    => y'/y = (x^2)(3ln(x) + 1) ...........simplify a bit
    => y' = y[(x^2)(3ln(x) + 1)] ..........solve for y'
    => y' = [x^(x^3)][(x^2)(3ln(x) + 1)]....write y' in terms of x

    2) y=(sin x)^(x^2) ......................same procedure
    => ln(y) = ln[(sin x)^(x^2)]
    => ln(y) = (x^2)ln(sin(x))
    => y'/y = 2xln(sin(x)) + (x^2)cos(x)/sin(x)
    => y' = y[2xln(sin(x)) + (x^2)cos(x)/sin(x)]
    => y' = [(sin x)^(x^2)][2xln(sin(x)) + (x^2)cos(x)/sin(x)]
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post
    Find dy/dx implicitly:
    1) 4xy+ln(x^2*y)=7
    1) 4xy + ln(x^2*y) = 7
    => 4y + 4x y' + [1/(x^2 *y)]*(2xy + x^2 y') = 0
    => 4x y' + [x^2/(x^2 *y)]y' = -4y - 2xy/(x^2 *y)
    => y' {4x + [x^2/(x^2 *y)]} = -4y - 2xy/(x^2 *y)
    => y' = [-4y - 2xy/(x^2 *y)]/{4x + [x^2/(x^2 *y)]}
    => dy/dx = [-4y - 2xy/(x^2 *y)]/{4x + [x^2/(x^2 *y)]}

    and you can simplify that a bit if you want
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    Hello, cupcakelova87!

    Find dy/dx implicitly: .4xy + ln(xy) .= .7
    I would simplify that log first . . .

    4xy + ln(x) + ln(y) .= .7 . . . .4xy + 2ln(x) + ln(y) .= .7


    Differentiate: .4xy' + 4y + 2/x + y'/y .= .0

    Then we have: .4xy' + y'/y .= .-4y - 2/x

    Factor: .(4x + 1/y)y' .= .-2(2y + 1/x)

    . . . . . . . . . . . 4xy + 1 . . . . . -2(2xy + 1)
    And we have: . --------- y' .= .--------------
    . . . . . . . . . . . . . .y . . . . . . . . . .x

    . . . . . . . . . . . . . .-2y(2xy + 1)
    Therefore: .y' . = . ---------------
    . . . . . . - . . . . . . . x(4xy + 1)


    I messed up my signs.
    They're correct now (I hope).
    Last edited by Soroban; April 6th 2007 at 09:11 AM.
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    Quote Originally Posted by Soroban View Post
    Hello, cupcakelova87!

    I would simplify that log first . . .

    4xy + ln(x) + ln(y) .= .7 . . . .4xy + 2ln(x) + ln(y) .= .7


    Differentiate: .4xy' + 4y + 2/x + y'/y .= .0

    Then we have: .4xy' + y'/y .= .-4y - 2/x

    Factor: .(4x + 1/y)y' .= .2(1/x - 2y)

    . . . . . . . . . . . 4xy + 1 . . . . . 2(1 - 2xy)
    And we have: . --------- y' .= .-----------
    . . . . . . . . . . . . . .y . . . . . . . . . .x

    . . . . . . . . . . . . . .2y(1 - 2xy)
    Therefore: .y' . = . --------------
    . . . . . . - . - . . . . .x(4xy + 1)

    yeah, usually i'd do that too, but i'm all scatter-brain today for some reason. thanks for looking out
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, cupcakelova87!
    Then we have: .4xy' + y'/y .= .-4y - 2/x

    Factor: .(4x + 1/y)y' .= .2(1/x - 2y)
    Oops... you forgot a minus sign there...
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