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.


Answer and Explanation: 1


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}.


Learn more about this topic:

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Partial Derivative: Definition, Rules & Examples

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Chapter 18 / Lesson 12
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What is a Partial Derivative? Learn to define first and second order partial derivatives. Learn the rules and formula for partial derivatives. See examples.


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