Find the indicated partial derivatives of: z = \frac{6x}{\sqrt{x^2 + y^2}} Find the indicated...

Question:

Find the indicated partial derivatives of:

{eq}z = \frac{6x}{\sqrt{x^2 + y^2}} {/eq}

Find the indicated partial derivative:

{eq}\frac{\partial }{\partial y} (\frac{y}{y + 5x}){/eq}

Quotient Rule and Parial Derivative:

Let {eq}k(x) {/eq} and {eq}m(x) {/eq} are the function of {eq}x {/eq} then the derivative of {eq}\frac{{k(x)}}{{m(x)}} {/eq} is given by:

{eq}{\,\frac{d}{{dx}}\frac{k}{m} = \frac{{m\frac{{dk}}{{dx}} - k\frac{{dm}}{{dx}}}}{{{m^2}}}} {/eq}

Let {eq}h(x, y) {/eq} be a function of two independent variables, then the partial derivative of a function of two variables can be found by applying the usual rules of differentiation with one exception. Since we have two variables, we must treat one as a constant when we take the derivative of the function with respect to the other variable

Answer and Explanation:

{eq}\displaystyle \eqalign{ & z = \frac{{6x}}{{\sqrt {{x^2} + {y^2}} }} \cr & {\text{Differentiating }}z\,\,{\text{w}}{\text{.r}}{\text{.t}}{\text{. '}}x{\text{';}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\frac{{6x}}{{\sqrt {{x^2} + {y^2}} }} \cr & \,\,\,\,\,\,\,\, = 6\frac{\partial }{{\partial x}}\frac{x}{{\sqrt {{x^2} + {y^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{d}{{dx}}cf(x) = x\frac{d}{{dx}}f(x)} \right) \cr & \,\,\,\,\,\,\,\, = 6\frac{{\sqrt {{x^2} + {y^2}} \frac{\partial }{{\partial x}}x - x\frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} }}{{\left( {{x^2} + {y^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From quotient rule}}} \right) \cr & \,\,\,\,\,\,\,\, = 6\frac{{\sqrt {{x^2} + {y^2}} - x\frac{x}{{\sqrt {{x^2} + {y^2}} }}}}{{\left( {{x^2} + {y^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right) \cr & \,\,\,\,\,\,\,\,\, = 6\frac{{{x^2} + {y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} \cr & \frac{{\partial z}}{{\partial x}} = 6\frac{{{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} \cr & \cr & {\text{Differentiating }}z\,\,{\text{w}}{\text{.r}}{\text{.t}}{\text{. '}}y{\text{';}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\frac{{6x}}{{\sqrt {{x^2} + {y^2}} }} \cr & \,\,\,\,\,\,\,\, = 6\frac{\partial }{{\partial y}}\frac{x}{{\sqrt {{x^2} + {y^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{d}{{dx}}cf(x) = x\frac{d}{{dx}}f(x)} \right) \cr & \,\,\,\,\,\,\,\, = 6\frac{{\sqrt {{x^2} + {y^2}} \frac{\partial }{{\partial y}}x - x\frac{\partial }{{\partial y}}\sqrt {{x^2} + {y^2}} }}{{\left( {{x^2} + {y^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From quotient rule}}} \right) \cr & \,\,\,\,\,\,\,\, = 6\frac{{0 - x\frac{x}{{\sqrt {{x^2} + {y^2}} }}}}{{\left( {{x^2} + {y^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right) \cr & \,\,\,\,\,\,\,\,\, = - 6\frac{{{x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} \cr & \frac{{\partial z}}{{\partial y}} = - 6\frac{{{x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} \cr & \cr & \frac{\partial }{{\partial y}}\left( {\frac{y}{{y + 5x}}} \right) \cr & \left( {\frac{{\left( {y + 5x} \right)\frac{\partial }{{\partial y}}y - y\frac{\partial }{{\partial y}}\left( {y + 5x} \right)}}{{{{\left( {y + 5x} \right)}^2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{From quotient rule}}} \right) \cr & \left( {\frac{{\left( {y + 5x} \right) - y\left( {1 + 0} \right)}}{{{{\left( {y + 5x} \right)}^2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{\partial }{{\partial y}}y = 1,\frac{\partial }{{\partial y}}x = 0} \right) \cr & \left( {\frac{{\left( {y + 5x - y} \right)}}{{{{\left( {y + 5x} \right)}^2}}}} \right) \cr & \frac{{5x}}{{{{\left( {y + 5x} \right)}^2}}} \cr} {/eq}


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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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