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Math Help - Implicit & Logarithmic differentiation problems.

  1. #1
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    Cool Implicit & Logarithmic differentiation problems.

    #1. Find the derivative of: e^x^y=sin(y^2)

    lne^x^y=lnsin(y^2)
    xylne=lnsin(y^2)
    xy*1=lnsin(y^2)
    y+xy'=lnsin(y^2)
    I need to use 2 chain rules on lnsin(y^2)
    y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]
    y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}
    y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'
    y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]
    y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}

    I'm sure there is something wrong in there



    #2. Find the derivative using logartihmic differentiation:
    y=\frac{x(x+1)^3}{(3x+1)^2}
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
    lny=[ln(x(x+1)^3]-[ln(3x+1)^2]
    Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.
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  2. #2
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    Re: Implicit & Logarithmic differentiation problems.

    Quote Originally Posted by jamesrb View Post
    [B]...



    #2. Find the derivative using logartihmic differentiation:
    y=\frac{x(x+1)^3}{(3x+1)^2}
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
    lny=[ln(x(x+1)^3]-[ln(3x+1)^2]
    Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.
    You were on the right track:

    lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}

    Now the differentiation of the RHS is quite simple.
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  3. #3
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    Re: Implicit & Logarithmic differentiation problems.

    Quote Originally Posted by earboth View Post
    You were on the right track:

    lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}

    Now the differentiation of the RHS is quite simple.
    Thanks, I think I got it:
    y'=y(\frac{1}{x}+\frac{3}{x+1}-\frac{6}{3x+1})
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  4. #4
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    Re: Implicit & Logarithmic differentiation problems.

    Quote Originally Posted by jamesrb View Post
    #1. Find the derivative of: e^x^y=sin(y^2)

    lne^x^y=lnsin(y^2)
    xylne=lnsin(y^2)
    xy*1=lnsin(y^2)
    y+xy'=lnsin(y^2)
    I need to use 2 chain rules on lnsin(y^2)
    y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]
    y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}
    y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'
    y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]
    y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}

    I'm sure there is something wrong in there



    #2. Find the derivative using logartihmic differentiation:
    y=\frac{x(x+1)^3}{(3x+1)^2}
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
    lny=[ln(x(x+1)^3]-[ln(3x+1)^2]
    Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
    lny=ln\frac{x(x+1)^3}{(3x+1)^2}
    You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.
    I think in Question 1, attempting logarithmic differentiation is a waste of time.

    \displaystyle \begin{align*} e^{x\,y} &= \sin{\left( y^2 \right)} \\ \frac{d}{dx} \left( e^{x\,y} \right) &= \frac{d}{dx}\left[ \sin{\left( y^2 \right)} \right] \\ e^{x\,y} \, \frac{d}{dx} \left( x\, y\right) &= \frac{d}{dy}\left[ \sin{\left( y^2 \right)} \right] \frac{dy}{dx} \\ e^{x\,y} \left( x\,\frac{dy}{dx} + y \right) &= 2y \cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ x\, e^{x\,y} \, \frac{dy}{dx} + y \, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} - x\, e^{x\,y} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= \left[ 2y\cos{\left( y^2 \right)} - x\, e^{x\,y} \right] \frac{dy}{dx} \\ \frac{y\, e^{x\,y}}{2y\cos{\left( y^2 \right)} - x \, e^{x\,y} } &= \frac{dy}{dx} \end{align*}
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