Find $\displaystyle \frac{ \partial z }{ \partial y } $ where $\displaystyle z$ is defined implicitly as a function of $\displaystyle x$ and $\displaystyle y$ by:
$\displaystyle sin(xyz) = x + 2y + 3z$
Thanks in advance
$\displaystyle \frac{\partial}{\partial y}\ [sin(xyz) = x+2y+3z]$
$\displaystyle cos(xyz)xz \frac{\partial z}{\partial y}\ = 0+2+3\frac{\partial z}{\partial y}\$
$\displaystyle \frac{\partial z}{\partial y}\ (xzcos(xyz)-3) = 2$
$\displaystyle \frac{\partial z}{\partial y}\ = \frac{2}{xzcos(xyz)-3}$
Could someone please check my work?
If z is an implicit function of x and y then the partial derivative of xyz with respect to y is $\displaystyle xz + xy \frac{\partial z}{\partial y}$.
Therefore the left hand side of the line marked * should be $\displaystyle \cos (xyz) \left(xz + xy \frac{\partial z}{\partial y} \right)$.