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

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

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

    c) $\displaystyle 4ln(4xyz) + sin(xz^2))$
    $\displaystyle F_x=\frac{4}{x}+z^2cos(xz^2)$
    $\displaystyle F_y=\frac{4}{y}$
    $\displaystyle F_z=\frac{4}{z}+2xzcos(xz^2)$
    *$\displaystyle \frac{dz}{dx}=-\frac{F_x}{F_z}=-\frac{\frac{4}{x}+z^2cos(xz^2)}{\frac{4}{z}+2xzcos (xz^2)}$
    *$\displaystyle \frac{dz}{dy}=-\frac{F_y}{F_z}=-\frac{\frac{4}{y}}{\frac{4}{z}+2xzcos(xz^2)}$
    *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
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    $\displaystyle xy + yz - xz = 0$

    $\displaystyle xy = -yz + xz $

    $\displaystyle xy = z(x-y ) $

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

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

    Therefore

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

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