Question

Differentiate implicitly to find the first partial derivatives of \(z\) \(x^{2}+2 y z+z^{2}=1\)

Short answer

The first derivative of \(z\) with respect to \(x\) is \(\frac{-x}{y+z}\), with respect to \(y\) is \(\frac{-z}{x+z}\), and with respect to \(z\) is 1.

Step-by-step solution

Differentiate with respect to \(x\)

Since the goal is to find partial derivatives of \(z\), the equation will first be differentiated with respect to \(x\): \(\frac{\partial }{\partial x} (x^{2}) + 2y\frac{\partial z}{\partial x} + 2z\frac{\partial z}{\partial x} = 0\). This simplifies to \(2x+2y\frac{\partial z}{\partial x} + 2z\frac{\partial z}{\partial x} = 0\). Solving for \(\frac{\partial z}{\partial x}\), we get \(\frac{\partial z}{\partial x} = \frac{-x}{y+z}\)

Differentiate with respect to \(y\)

The equation is then differentiated with respect to \(y\): \(2x\frac{\partial z}{\partial y} + 2z+2z\frac{\partial z}{\partial y}= 0\). This simplifies to \(2x\frac{\partial z}{\partial y} + 2z\frac{\partial z}{\partial y} + 2z = 0\). Solving for \(\frac{\partial z}{\partial y}\), we get \(\frac{\partial z}{\partial y} = \frac{-z}{x+z}\)

Differentiate with respect to \(z\)

Lastly, the equation is differentiated with respect to \(z\): \(2x\frac{\partial z}{\partial z} + 2y+2z= 0\). This simplifies to \(2x+2y+2z = 0\). Therefore, the first partial derivative of \(z\) with respect to \(z\), \(\frac{\partial z}{\partial z}\) is equal to 1.