Q. 32 - Calculus

Question

Find the first-order partial derivatives for the functions in Exercises 27–36.  

$g\left(x,y\right)=x\cos y$

Step-by-Step Solution

Answer

The first-order partial derivatives are $\frac{\partial g}{\partial x}=\cos y \text{and} \frac{\partial g}{\partial y}=-x\sin y$

Step 1: Given

The given function is: $g\left(x,y\right)=x\cos y$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

$$\begin{gathered} g\left(x,y\right)=x\cos y \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left(x\cos y\right) \\ \frac{\partial g}{\partial x}=\cos y \\ g\left(x,y\right)=x\cos y \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left(x\cos y\right) \\ \frac{\partial g}{\partial y}=x\left(-\sin y\right) \\ \frac{\partial g}{\partial y}=-x\sin y \\ \text{Hence}, \frac{\partial g}{\partial x}=\cos y \text{and} \frac{\partial g}{\partial y}=-x\sin y \end{gathered}$$