Q. 27 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The first order partial derivatives are $f_x\left(x,y\right)=e^x\sin \left(xy\right)+y{e}^x\cos \left(xy\right) \text{and} f_y\left(x,y\right)=x{e}^x\cos \left(xy\right)$

Step 1: Given

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

Step 2: To find

We have to find the first-order partial derivatives.

Step 3: Calculation

$$\begin{gathered} f\left(x,y\right)=e^x\sin \left(xy\right) \\ {f}_x\left(x,y\right)=\frac{\partial }{\partial x}\left(e^x\sin \left(xy\right)\right) \\ {f}_x\left(x,y\right)=\frac{\partial }{\partial x}\left(e^x\right)\sin \left(xy\right)+e^x\frac{\partial }{\partial x}\left(\sin \left(xy\right)\right) \\ {f}_x\left(x,y\right)=e^x\sin \left(xy\right)+e^x\cos \left(xy\right)\frac{\partial }{\partial x}\left(xy\right) \\ {f}_x\left(x,y\right)=e^x\sin \left(xy\right)+e^x\cos \left(xy\right)\left(y\right) \\ {f}_x\left(x,y\right)=e^x\sin \left(xy\right)+y{e}^x\cos \left(xy\right) \\ f\left(x,y\right)=e^x\sin \left(xy\right) \\ {f}_y\left(x,y\right)=\frac{\partial }{\partial y}\left(e^x\sin \left(xy\right)\right) \\ {f}_y\left(x,y\right)=\frac{\partial }{\partial y}\left(e^x\right)\sin \left(xy\right)+e^x\frac{\partial }{\partial y}\left(\sin \left(xy\right)\right) \\ {f}_y\left(x,y\right)=\left(0\right)\sin \left(xy\right)+e^x\left(\cos \left(xy\right)\right)\frac{\partial }{\partial y}\left(xy\right) \\ {f}_y\left(x,y\right)=0+e^x\cos \left(xy\right)\left(x\right) \\ {f}_y\left(x,y\right)=x{e}^x\cos \left(xy\right) \\ \text{Hence}, f_x\left(x,y\right)=e^x\sin \left(xy\right)+y{e}^x\cos \left(xy\right) \text{and} f_y\left(x,y\right)=x{e}^x\cos \left(xy\right) \end{gathered}$$