Question

Differentiate implicitly to find the first partial derivatives of \(z\) \(x \ln y+y^{2} z+z^{2}=8\)

Short answer

The first partial derivatives of \(z\) are \(\frac{\partial z}{\partial x}=\frac{-\ln y}{y^{2}+2z}\) and \(\frac{\partial z}{\partial y}=\frac{-\frac{x}{y}-2yz}{y^{2}+2z}\)

Step-by-step solution

Formulate the derivatives

We need to find the two first partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). To do that, we'll start by differentiating both sides of the equation with respect to \(x\) and \(y\) respectively.

Differentiate with respect to \(x\)

First, differentiate with respect to \(x\): \(\frac{\partial}{\partial x}(x \ln y+y^{2} z+z^{2})=\frac{\partial}{\partial x}(8)\). This gives: \(\ln y + y^{2} \frac{\partial z}{\partial x}+ 2z \frac{\partial z}{\partial x}=0\). Now, isolate \(\frac{\partial z}{\partial x}\) to get: \(\frac{\partial z}{\partial x}=\frac{-\ln y}{y^{2}+2z}\).

Differentiate with respect to \(y\)

Next, differentiate with respect to \(y\): \(\frac{\partial}{\partial y}(x \ln y+y^{2} z+z^{2})=\frac{\partial}{\partial y}(8)\). This results in: \(\frac{x}{y} + 2yz + y^{2} \frac{\partial z}{\partial y}+2z \frac{\partial z}{\partial y}=0\). Isolate \(\frac{\partial z}{\partial y}\) to get: \(\frac{\partial z}{\partial y}=\frac{-\frac{x}{y}-2yz}{y^{2}+2z}\)