For the following function, evaluate the partial derivatives below. F\left ( w, \ x, \ y, \ z...

Question:


For the following function, evaluate the partial derivatives below.

{eq}F\left ( w, \ x, \ y, \ z \right ) = 3xy^2 + \frac{w^3z^3}{32y} - \frac{xy^2z^3}{w} \\ a) \ \left ( \frac{\partial F}{\partial y} \right )_{w, \ x, \ z} \\ b) \ \left [ \frac{\partial }{\partial x}\left ( \frac{\partial F}{\partial z} \right )_{w, \ x, \ y} \right ]_{w, \ y, \ z} \\ c) \ \left \{ \frac{\partial }{\partial w}\left [ \frac{\partial }{\partial z}\left ( \frac{\partial F}{\partial x} \right )_{w, \ y, \ z} \right ]_{w, \ x, \ y} \right \}_{x, \ y, \ z} {/eq}

Partial derivative


The partial derivative is the derivative in which a function having more than one variable and the derivative of such function is find with respect to one variable by treating other variables as a constant. The application of partial derivative is in the field of differential calculus and vector calculus.

Answer and Explanation:


Given that,

{eq}F\left( {w,x,y,z} \right) = \left( {3xy^2 + \dfrac{{w^3 z^3 }}{{32y}} - \dfrac{{xy^2 z^3 }}{w}} \right) {/eq}

Solution (a) Partially differentiate with respect to {eq}y {/eq} , we get

{eq}\begin{align*} \left( {\dfrac{{\partial F}}{{\partial y}}} \right)_{w,x,z} &= \dfrac{\partial }{{\partial y}}\left( {3xy^2 + \dfrac{{w^3 z^3 }}{{32y}} - \dfrac{{xy^2z^3 }}{w} } \right) \\\\ &= \left( {3x\left( {2y} \right)} \right) + \left( {\left( { - \dfrac{1}{{y^2 }}} \right)} \right)\left( {\dfrac{{w^3 z^3 }}{{32}}} \right) - \left( {\dfrac{{2xyz^3 }}{w}} \right) \\\\ &= \left( {6xy} \right) - \left( {\dfrac{{w^3 z^3 }}{{32y^2 }}} \right) - \left( {\dfrac{{2xyz^3 }}{w}} \right) \\\\ \end{align*} {/eq}

Hence, this is our required solution.

Solution (b) According to question,

{eq}\begin{align*} \left[ {\dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial F}}{{\partial z}}} \right)_{w,x,y} } \right]_{w,y,z} &= \dfrac{\partial }{{\partial x}}\left[ {\dfrac{\partial }{{\partial z}}\left( {3xy^2 + \dfrac{{w^3 z^3 }}{{32y}} - \dfrac{{xy^2 z^3 }}{w}} \right)} \right] \\\\ &= \dfrac{\partial }{{\partial x}}\left[ {0 + \left( {\dfrac{{3w^3 z^2 }}{{32y}}} \right) - \left( {\dfrac{{3xy^2 z^2 }}{w}} \right)} \right] \\\\ &= \left[ {0 + 0 - \left( {\dfrac{{3y^2 z^2 }}{w}} \right)} \right] \\\\ &= \left[ { - \dfrac{{3y^2 z^2 }}{w}} \right] \\\\ \end{align*} {/eq}

Hence, this is our required solution.

Solution (c) According to question,

{eq}\begin{align*} \left\{ {\dfrac{\partial }{{\partial w}}\left[ {\dfrac{\partial }{{\partial z}}\left( {\dfrac{{\partial F}}{{\partial x}}} \right)_{w,y,z} } \right]_{w,x,y} } \right\}_{x,y,z} &= \dfrac{\partial }{{\partial w}}\left[ {\dfrac{\partial }{{\partial z}}\left\{ {\dfrac{\partial }{{\partial x}}\left( {3xy^2 + \dfrac{{w^3 z^3 }}{{32y}} - \dfrac{{xy^2 z^3 }}{w}} \right)} \right\}} \right] \\\\ &= \dfrac{\partial }{{\partial w}}\left[ {\dfrac{\partial }{{\partial z}}\left\{ {3y^2 + 0 - \dfrac{{y^2 z^3 }}{w}} \right\}} \right] \\\\ &= \dfrac{\partial }{{\partial w}}\left[ {0 - \dfrac{{3y^2 z^2 }}{w}} \right] \\\\ &= \left[ {\dfrac{{3y^2 z^2 }}{{w^2 }}} \right] \\\\ \end{align*} {/eq}

Hence, this is our required solution.


Learn more about this topic:

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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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