# Find {eq}\displaystyle f_x {/eq} for the function {eq}\displaystyle f(x,y) = \frac{x+y}{(xy-1)} {/eq}.

## Question:

Find {eq}\displaystyle f_x {/eq} for the function {eq}\displaystyle f(x,y) = \frac{x+y}{(xy-1)} {/eq}.

## Partial Derivative:

We can find the partial derivative of a function with respect {eq}x{/eq} by differentiating the function for {eq}x{/eq} and treating {eq}y{/eq} as a constant throughout the process. We proceed with normal differentiation, and the partial derivation is performed for multivariable function for single variable function. The rules for partial derivation are similar to normal differentiation.

Given:

• The function is {eq}f\left( {x,y} \right) = \frac{{x + y}}{{\left( {xy - 1} \right)}}{/eq}.

To find {eq}{f_x}\left( {x,y} \right){/eq}, take partial differentiation of {eq}f\left( {x,y} \right){/eq} with respect to {eq}x{/eq}, to get:

{eq}\frac{\partial }{{\partial x}}f\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {\frac{{x + y}}{{\left( {xy - 1} \right)}}} \right){/eq}

Using quotient rule to get:

{eq}\begin{align*} {f_x}\left( {x,y} \right) &= \frac{{\left( {xy - 1} \right)\frac{\partial }{{\partial x}}\left( {x + y} \right) - \left( {x + y} \right)\frac{\partial }{{\partial x}}\left( {xy - 1} \right)}}{{{{\left( {xy - 1} \right)}^2}}}\\ &= \frac{{\left( {xy - 1} \right)\left( {\frac{\partial }{{\partial x}}x + \frac{\partial }{{\partial x}}y} \right) - \left( {x + y} \right)\left( {y\frac{\partial }{{\partial x}}x - \frac{\partial }{{\partial x}}1} \right)}}{{{{\left( {xy - 1} \right)}^2}}}\\ &= \frac{{\left( {xy - 1} \right)\left( {1 + 0} \right) - \left( {x + y} \right)\left( {y \times 1 - 0} \right)}}{{{{\left( {xy - 1} \right)}^2}}}\\ &= \frac{{\left( {xy - 1} \right) - y\left( {x + y} \right)}}{{{{\left( {xy - 1} \right)}^2}}} \end{align*}{/eq}

Solving further,

{eq}\begin{align*} {f_x}\left( {x,y} \right) &= \frac{{xy - 1 - xy - {y^2}}}{{{{\left( {xy - 1} \right)}^2}}}\\ &= \frac{{ - {y^2} - 1}}{{{{\left( {xy - 1} \right)}^2}}} \end{align*}{/eq}

The partial derivative of the function is {eq}{f_x}\left( {x,y} \right) = - \frac{{({y^2} + 1)}}{{{{\left( {xy - 1} \right)}^2}}}{/eq}. 