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Math Help - Derivative question

  1. #1
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    Derivative question

    Hi everyone

    Need help to verify this working.Thank you in advance for all help & support,really appreciate.

    Find \frac{dz}{dx}& \frac{dz}{dy} for
    a) xy + yz - xz = 0
    y + y\frac{dz}{dx}-z-x\frac{dz}{dx}=0
    \frac{dz}{dx}(y-x)=z-y
    \frac{dz}{dx}=\frac{z-y}{y-x}

    x+z+y\frac{dz}{dy}-x\frac{dz}{dy}=0
    \frac{dz}{dy}(y-x)=-x-z
    \frac{dz}{dy}=\frac{-x-z}{y-x}

    b) xyz-cos(x+y+z)
    F_x =yz+sin(x+y+z)
    F_y=xz+sin(x+y+z)
    F_z= xy+sin(x+y+z)
    \frac{dz}{dx}=-\frac{F_x}{F_z}=-\frac{yz+sin(x+y+z)}{xy+sin(x+y+z)}
    \frac{dz}{dy}=-\frac{F_y}{F_z}=-\frac{xz+sin(x+y+z)}{xy+sin(x+y+z)}

    c) 4ln(4xyz) + sin(xz^2))
    F_x=\frac{4}{x}+z^2cos(xz^2)
    F_y=\frac{4}{y}
    F_z=\frac{4}{z}+2xzcos(xz^2)
    * \frac{dz}{dx}=-\frac{F_x}{F_z}=-\frac{\frac{4}{x}+z^2cos(xz^2)}{\frac{4}{z}+2xzcos  (xz^2)}
    * \frac{dz}{dy}=-\frac{F_y}{F_z}=-\frac{\frac{4}{y}}{\frac{4}{z}+2xzcos(xz^2)}
    *Not sure how to simplify both equations,just left in this form.
    Thank you in advance for all your kind help & support, really hope someone can verify for me.
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  2. #2
    Master Of Puppets
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     xy + yz - xz = 0

     xy = -yz + xz

     xy = z(x-y )

     \frac{xy}{x-y } = z

     z=\frac{xy}{x-y }

    Therefore

    \frac{dz}{dx}= \frac{(x-y)y- xy}{(x-y)^2 }=\dots

     \frac{dz}{dy}=\frac{(x-y)x- xy}{(x-y)^2 }=\dots
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