91Ó°ÊÓ

Introduction:

In this question it is provide equation $u=e^x\cos y$and to verify that

${∂}^{2}u/∂x∂y={∂}^{2}u/∂y∂x$ and ${∂}^{2}u/∂x^2={∂}^{2}u/∂y^2=0$.

The procedure of calculating the partial derivative of a function is called partial differentiation. This method is used to get the partial derivative of a function with respect to one variable while keeping the other constant.

(a)

Consider that,

$u=e^x\cos y$

Implies that,

role="math" localid="1659092296034" $\frac{∂u}{∂x}=e^x\cos y$

Implies that,

$\frac{∂}{∂y}\left(\frac{∂u}{∂x}\right)=-e^x\sin y$

Implies that,

$\frac{{∂}^{2}u}{∂x∂y}=-e^x\sin y\_\_\_\_\_\_\left(1\right)$

Now,

$\frac{∂u}{∂y}=-e^x\sin y$

Since,

$\frac{{∂}^{2}u}{∂y∂x}=-e^x\sin y\_ \_ \_ \_ \_\left(2\right)$

(b)

Consider $u=e^x\cos y$

Impleis that,

$\frac{∂u}{∂x}=-e^x\cos y$

Implies that,

$\frac{{∂}^{2}u}{∂x^2}=-e^x\cos y\_ \_ \_ \_ \_ \left(3\right)$

And,

role="math" localid="1659094574208" $\frac{{∂}^{2}u}{∂y∂x}=-e^x\cos y\_ \_ \_ \_ \_ \left(4\right)$

Add equation (3) and (4) ,

$\frac{{∂}^{2}u}{∂x^2}+\frac{{∂}^{2}u}{∂y^2}=-e^x\cos y-e^x\cos y\_ \_ \_ \_ \_ \left(4\right)$

Implies that,

$\frac{{∂}^{2}u}{∂x^2}+\frac{{∂}^{2}u}{∂y^2}=0$