# Partial Derivative of xy [With Respect to x and y]

The partial derivative of xy with respect to x is equal to y and the partial derivative of xy with respect to y is equal to x. Their formula is given below:

1. $\dfrac{\partial }{\partial x}(xy)=y$
2. $\dfrac{\partial }{\partial y}(xy)=x$

## Partial Derivative of xy with respect to x

Let us will find the partial derivative of xy using the definition.

Let f(x, y) = xy.

By definition of partial derivatives, we have

$\dfrac{\partial }{\partial x}(xy)$ = limh→0 $\dfrac{f(x+h, y)- f(x,y)}{h}$

= limh→0 $\dfrac{(x+h)y- xy}{h}$

= limh→0 $\dfrac{xy+hy- xy}{h}$

= limh→0 $\dfrac{hy}{h}$

= limh→0 y

= y

So the partial derivative of xy with respect to x is y and this is obtained by the definition of partial derivatives.

Question 1: If z=xy, then find ∂z/∂x.

To find ∂z/∂x, that is, the partial derivative of xy with respect to x, we will treat y as a constant and x as a variable. That is,

∂z/∂x = $\dfrac{\partial }{\partial x}(xy)$ = $y \dfrac{d}{dx}(x)$ = $y$.

Thus, if z=xy then ∂z/∂x=y.

## Partial Derivative of xy with respect to y

By the definition of partial derivatives, we have

$\dfrac{\partial }{\partial y}(xy)$ = limk→0 $\dfrac{f(x, y+k)- f(x,y)}{k}$ where f(x, y) = xy.

= limk→0 $\dfrac{x(y+k)- xy}{k}$

= limk→0 $\dfrac{xy+xk- xy}{k}$

= limk→0 $\dfrac{xk}{k}$

= limk→0 x

= x

So the partial derivative of xy with respect to y is x and this is obtained by the definition of partial derivatives.

Question 2: If z=xy, then find ∂z/∂y.

While finding ∂z/∂y, that is, the partial derivative of xy with respect to y, we will treat x as a constant and y as a variable. So we have

∂z/∂y = $\dfrac{\partial }{\partial y}(xy)$ = $x \dfrac{d}{dy}(y)$ = $x$.

Therefore, if z=xy then ∂z/∂y = x.

## FAQs

### Q1: What is the partial derivative of xy with respect to x?

Answer: The partial derivative of xy with respect to x is equal to y.

### Q2: What is the partial derivative of xy with respect to y?

Answer: The partial derivative of xy with respect to y is equal to x.