Question

Differentiate implicitly to find the first partial derivatives of \(z\) \(e^{x z}+x y=0\)

Short answer

\(\frac{\partial z}{\partial x}= -\frac{y}{z+x}\), \(\frac{\partial z}{\partial y} = -\frac{1}{e^{xz}}\)

Step-by-step solution

Differentiate With Respect to x

First, differentiate the entire equation with respect to \(x\), treating \(y\) and \(z\) as functions of \(x\). The derivative of \(e^{x z}\) can be found using the chain rule as \(e^{x z} (z+ x \frac{\partial z}{\partial x})\). The derivative of \(x y\) is \(y + x \frac{\partial y}{\partial x}\), according to the product rule. So, the differentiated equation becomes: \(e^{x z} (z+ x \frac{\partial z}{\partial x}) + y + x \frac{\partial y}{\partial x} =0\).

Differentiate With Respect to y

Perform the similar differentiation with respect to \(y\). The derivative of \(e^{x z}\) will now be \(x e^{x z} \frac{\partial z}{\partial y}\) using chain rule and the derivative of \(x y\) will just be \(x\). Hence, the differentiated equation will be: \(x e^{x z} \frac{\partial z}{\partial y} + x = 0\).

Solve For Required Derivatives

From the first equation, isolate \(\frac{\partial z}{\partial x}\), get: \(\frac{\partial z}{\partial x}= -\frac{y}{z+x}\). From the second equation, isolate \(\frac{\partial z}{\partial y}\), get: \(\frac{\partial z}{\partial y} = -\frac{1}{e^{xz}}\). These are the required first partial derivatives.