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Basic partial derivatives : u = $x^2y$ + $e^{xy^2}$, Show that $ \frac{\partial^2 \; u}{\partial x \partial y} \; = \; \frac{\partial^2 \; u}{\partial y \partial x} $
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Now, $ \frac{\partial u}{\partial x} \; = \; \frac{\partial }{\partial x} \bigg( x^2y \; + \; e^{xy^2} \bigg) \; = \; \frac{\partial }{\partial x} (x^2y) \; + \; \frac{\partial }{\partial x} (e^{xy^2}) \\ \therefore \frac{\partial u}{\partial x} \; = \; 2xy + y^2e^{xy^2} \; \; \; \ldots (i) \\ \; \\ Similarly, \; \frac{\partial u}{\partial y} \; = \; \frac{\partial }{\partial y} \bigg( x^2y \; + \; e^{xy^2} \bigg) \; = \; \frac{\partial }{\partial y} (x^2y) \; + \; \frac{\partial }{\partial y} (e^{xy^2}) \\ = x^2 \; + \; e^{xy^2} \; \frac{\partial }{\partial y} xy^2 \\ \therefore \frac{\partial u }{\partial y} \; = \; x^2 + 2xye^{xy^2} \; \; \; \ldots (ii) \\ \; \\ \; \\ Now, \\ \frac{\partial^2 \; u}{\partial x \partial y} \; = \; \frac{\partial}{\partial x } \bigg( \frac{\partial u}{\partial y} \bigg) \; = \; \frac{\partial}{\partial x } \Big[ x^2 + 2xye^{xy^2} \Big] \; \; \; \; \; \ldots From \; (ii) \\ = \frac{\partial}{\partial x } (x^2) \; + \; \frac{\partial}{\partial x } (2xye^{xy^2}) \\ = \frac{\partial}{\partial x } (x^2) \; + \; 2xy \frac{\partial}{\partial x }(e^{xy^2}) \; + \; e^{xy^2} \frac{\partial}{\partial x } (2xy) \\ \therefore \frac{\partial^2 \; u}{\partial x \partial y} \; = \; 2x \; + \; 2xye^{xy^2} \frac{\partial}{\partial x } (xy^2) \; + \; e^{xy^2}2y \\ \therefore \frac{\partial^2 \; u}{\partial x \partial y} \; = \; 2x \; + \; 2xy.y^2e^{xy^2} + 2y.e^{xy^2} \\ \therefore \frac{\partial^2 \; u}{\partial x \partial y} \; = \; 2x + \; (2xy^3 \; + \; 2y)e^{xy^2} \; \; \; \; \ldots (iii) \\ \; \\ Similarly, \\ \frac{\partial^2 \; u}{\partial y \partial x} \; = \; \frac{\partial}{\partial y } \bigg( \frac{\partial u}{\partial x} \bigg) \; = \; \frac{\partial}{\partial y } \Big[ 2xy + y^2e^{xy^2} \Big] \; \; \; \; \; \ldots From \; (i) \\ = \frac{\partial}{\partial y } (2xy) \; + \; \frac{\partial}{\partial y } (y^2e^{xy^2}) \\ = \frac{\partial}{\partial y } (2xy) \; + \; y^2 \frac{\partial}{\partial y }(e^{xy^2}) \; + \; e^{xy^2} \frac{\partial}{\partial y } (y^2) \\ = \; 2x \; + \; y^2e^{xy^2} \frac{\partial}{\partial y } (xy^2) \; + \; e^{xy^2}2y \\ = \; 2x \; + \; y^2e^{xy^2}(2xy) + 2y.e^{xy^2} \\ \therefore \frac{\partial^2 \; u}{\partial y \partial x} \; = \; 2x + \; (2xy^3 \; + \; 2y)e^{xy^2} \; \; \; \; \ldots (iv) \\ \; \\ \; \\ From \; (iii) \; and \; (iv), \; it \; is \; proved \; that \; \frac{\partial^2 \; u}{\partial x \partial y} = \frac{\partial^2 \; u}{\partial y \partial x} $

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