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Thread: Partial derivative, multivariable

  1. #1
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    Partial derivative, multivariable

    $\displaystyle f(x,y)$

    For all s,t :
    $\displaystyle f(s,8s)=6s+cos(6s)$
    $\displaystyle f(t,-7t)=1+6t+5t^2$

    Determin $\displaystyle \frac{\partial f}{\partial x}(0,0) $ and $\displaystyle \frac{\partial f}{\partial y}(0,0) $

    Im new to multi variables so i donīt know how to do this.

    I thought if i set $\displaystyle z=f(x(s,t),y(s,t))$

    i get:

    $\displaystyle \frac{\partial z}{\partial s }= \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} $

    $\displaystyle \frac{\partial z}{\partial t }= \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} $

    from here i donīt know what to do.

    Thanks!
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  2. #2
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    Quote Originally Posted by mechaniac View Post
    $\displaystyle f(x,y)$

    For all s,t :
    $\displaystyle f(s,8s)=6s+cos(6s)$
    $\displaystyle f(t,-7t)=1+6t+5t^2$

    Determin $\displaystyle \frac{\partial f}{\partial x}(0,0) $ and $\displaystyle \frac{\partial f}{\partial y}(0,0) $

    Im new to multi variables so i donīt know how to do this.

    I thought if i set $\displaystyle z=f(x(s,t),y(s,t))$

    i get:

    $\displaystyle \frac{\partial z}{\partial s }= \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} $

    $\displaystyle \frac{\partial z}{\partial t }= \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} $

    from here i donīt know what to do.

    Thanks!
    You have a good start. Using the chain rule that you have above gives

    $\displaystyle \frac{\partial f}{\partial s}=6-6\sin(6s)=\frac{\partial f}{\partial x}\cdot 1+\frac{\partial f}{\partial y}\cdot 8 $

    Using the subscript notation for partial derivatives gives
    $\displaystyle f_x(s,8s)+8f_y(s,8s)=6-6\sin(6s)$

    $\displaystyle \frac{\partial f}{\partial t}=6+10t=\frac{\partial f}{\partial x}\cdot 1+\frac{\partial f}{\partial y}\cdot (-7) $

    $\displaystyle f_x(t,-7t)-7f_y(t,-7t)=6+10t$

    Now if
    $\displaystyle t=s=0$

    You will have a system of equations to find the partial derivatives you want!
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  3. #3
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    Thanks! i just got confused by all the variables
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