Q. 31 - Calculus

Question

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

$f\left(x,y\right)=x\sin y$

Step-by-Step Solution

Answer

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

Step 1: Given

The given function is:  $f\left(x,y\right)=x\sin y$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

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