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If $z=f(x,y)\;, x=e^u+e^{-v} \;, x=e^{-u}-e^{v}$prove that $\dfrac{\partial z}{\partial u} - \dfrac{\partial z}{\partial v}= x\dfrac{\partial z}{\partial x} - y\dfrac{\partial z}{\partial y}$
partial differentiation • 716  views
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x$=e^u+e^{-v} \; \; \& \; \; y=e^{-u}-e^v \; \; \; \ldots (i) \\ \; \\ \; \\ \dfrac{\partial x}{\partial u} \;=\; e^{v} \\ \dfrac{\partial y}{\partial u} \;=\; -e^{-u} \; \; \; \ldots (ii) \\ \; \\ \; \\ \dfrac{\partial x}{\partial v} \;=\; -e^{-v} \\ \dfrac{\partial x}{\partial v} \;=\; -e^v \; \; \; \ldots (iii)$

By chain rule,

$\therefore \dfrac{\partial z}{\partial u} \;=\; \dfrac{\partial z}{\partial x} . \dfrac{\partial x}{\partial u} + \dfrac{\partial z}{\partial y} . \dfrac{\partial y}{\partial u} \\ \; \\ \; \\ \therefore \dfrac{\partial z}{\partial u} \;=\; e^u . \dfrac{\partial z}{\partial x} - e^{-u} . \dfrac{\partial z}{\partial y} \; \; \; \ldots from (ii) \ldots (iv) \\ \; \\ \; \\ Also, \dfrac{\partial z}{\partial v} \;=\; \dfrac{\partial z}{\partial x} . \dfrac{\partial x}{\partial v} + \dfrac{\partial z}{\partial y} . \dfrac{\partial y}{\partial v} \\ \; \\ \; \\ \therefore \dfrac{\partial z}{\partial v} \;=\; -e^{-v} . \dfrac{\partial z}{\partial x} - -e^{v} . \dfrac{\partial z}{\partial y} \; \; \; \ldots from (iii) \ldots (v)$

Equation (iv) – (v) gives

$\therefore \dfrac{\partial z}{\partial u} - \dfrac{\partial z}{\partial v} \;=\; e^u . \dfrac{\partial z}{\partial x} - e^{-u} . \dfrac{\partial z}{\partial y} + e^{-v} . \dfrac{\partial z}{\partial x} + e^v . \dfrac{\partial z}{\partial y} \\ \; \\ \; \\ \therefore \dfrac{\partial z}{\partial u} - \dfrac{\partial z}{\partial v} \;=\; (e^u+e^{-v} ) . \dfrac{\partial z}{\partial x} - (e^{-u}+e^{v} ) . \dfrac{\partial z}{\partial y} + \\ \; \\ \; \\ \therefore \therefore \dfrac{\partial z}{\partial u} - \dfrac{\partial z}{\partial v} \;=\; x \dfrac{\partial z}{\partial x} - y \dfrac{\partial z}{\partial v} \; \; \; \; from \; (i) \; \; \; Hence \; Proved.$

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