Suppose z = z(x,y) is given implicitly by the equation xyz = (x+y) z. Find z x and ...


Suppose{eq}\displaystyle \ z = z(x,y) {/eq} is given implicitly by the equation{eq}\displaystyle \ \frac {xy}{z} = (x+y) \ln z. {/eq}

Find{eq}\displaystyle \ \frac {\partial z}{\partial x} {/eq} and{eq}\displaystyle \ \frac {\partial z}{\partial y}. {/eq}

Implicit Derivation:

The functions, ideally could be given in explicit form, but this is not always possible.

For the case of the implicit derivation, the partial derivatives are calculated, in two variables such as:

{eq}F\left( {x,y,z} \right) = 0 \to \left\{ \begin{array}{l} \frac{{\partial z}}{{\partial x}} = - \frac{{\frac{{\partial F}}{{\partial x}}}}{{\frac{{\partial F}}{{\partial z}}}}\\ \frac{{\partial z}}{{\partial y}} = - \frac{{\frac{{\partial F}}{{\partial y}}}}{{\frac{{\partial F}}{{\partial z}}}} \end{array} \right. {/eq}

Answer and Explanation: 1

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From the equation, we extract to the function with which we will calculate the partial derivatives:

{eq}\frac{{xy}}{z} = (x + y)\ln z\\ \;F\left(...

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Learn more about this topic:

Partial Derivative: Definition, Rules & Examples


Chapter 18 / Lesson 12

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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