# Math Help - Partial Derivatives

1. ## Partial Derivatives

Could someone show me where to go at ??? please

dz/dx is representing partial z with respect to x.

Find the value of dz/dx at the point (1,1,1) if the equation

xy+(z^3)x-2yz=0

defines z as a function of the two independent variables x and y and the partial derivative.

so we need to differentiate with respect to x?

d/dx(xz) + d/dx((z^3)x)-d/dx(2yz) = 0

x dz/dx + ??? - 2y dz/dx = 0

2. Originally Posted by ur5pointos2slo
Could someone show me where to go at ??? please

dz/dx is representing partial z with respect to x.

Find the value of dz/dx at the point (1,1,1) if the equation

xy+(z^3)x-2yz=0

defines z as a function of the two independent variables x and y and the partial derivative.

so we need to differentiate with respect to x?

d/dx(xz) + d/dx((z^3)x)-d/dx(2yz) = 0

x dz/dx + ??? - 2y dz/dx = 0
Yes, you partially differentiate both sides of the equation with respect to x.

$\frac{\partial}{\partial x}\left[xy+z^3x-2yz\right]=\frac{\partial}{\partial x}0\implies y+\frac{\partial}{\partial x}\left(z^3x\right)-2\frac{\partial}{\partial x}\left[yz\right]=0$

Now since $z=z\!\left(x,y\right)$, we use the product rule to evaluate $\frac{\partial}{\partial x}\left[z^3x\right]$:

$\frac{\partial}{\partial x}\left[z^3x\right]=z^3+2z^2x\frac{\partial z}{\partial x}$

In the case of $\frac{\partial}{\partial x}\left[yz\right]$, we just have $y\frac{\partial z}{\partial x}$ since y is a constant and z depends on x.

Thus, it follows that we have $y+z^3+2z^2x\frac{\partial z}{\partial x}-2y\frac{\partial z}{\partial x}=0\implies y+z^3=\left(2y-2z^2x\right)\frac{\partial z}{\partial x}\implies \frac{\partial z}{\partial x}=\frac{y+z^3}{2y-2z^2x}$

Does this make sense?

3. That makes sense. Thank you.