by Integr88 » Thu 24 March, 2011 07:20
The question says find `f_(x,x,y,z)` of the funciton `f(x,y,z)=sin(3x+yz)`
it means take the derivatives of `sin(3x+yz)` as in the following order:
`f_x` take the derivative with respect of x
`f_(x\x)` take the derivative of the step above `f_x` with respect to x again. so it is `f_(x\x)`
`f_(x\x\y)` now take the derivatives of the result of `f_(x\x)` with respect to y
and then
`f_(x\x\y\z)` and take the derivatives of the one step above with respect to z
taking `f_x` we have `sin(3x+yz)`
for `f_x=cos(3x-yz)*3x` or `3xcos(3x-yz)`
taking `f_(x\x)` we have `3xcos(3x-yz)` apply product rule and take the derivatives with respect to x
`f_(x\x)=3*cos(3x-yz)+3x*(-sin(3x-yz)*3x)=3cos(3x-yz)-9xsin(3x-yz)`
taking `f_(x\x\y)` we have now `3cos(3x-yz)-9xsin(3x-yz)` so take the derivatives with respect to y
`f_(x\x\y)=[0*cos(3x-yz)+3*(-sin(3x-yz))*(z) ]=0-3zsin(3x-yz)=-3zsin(3x-yz)`
taking `f_(x\x\y\z)` we have `-3zsin(3x-yz)` take the derivative with respect to z
`f_(x\x\y\z)=-3cos(3x-yz)+(-3cos(3x-yz)*(-y))=-3sin(3x-yz)+3ycos(3x-yz)`
`:.` the partial derivatives `f_(x,x,y,z)`of the function `sin(3x+yz)` is `-3sin(3x-yz)+3ycos(3x-yz)`