if u=e^(xyz) then u_(xyx) =? a)u((xyz)^2 +3xyz+1) b)u(3(xyz)^2 +1) c)u((xyz)^2 +2yz+1) please help


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Question Number 53378 by Necxx last updated on 21/Jan/19

$i f u = e^{x y z} t h e n u_{x y x} = ?$ $a) u ({(x y z)}^{2} + 3 x y z + 1) b) u (3 {(x y z)}^{2} + 1)$ $c) u ({(x y z)}^{2} + 2 y z + 1)$ $p l e a s e h e l p$
	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

	$u = e^{x y z}$ $u_{x} = e^{x y z} \times \frac{\partial}{\partial x} (x y z) = {y z e}^{x y x}$ $u_{x y} = \frac{\partial}{\partial y} ({y z e}^{x y z}) = e^{x y z} \times z \frac{\partial}{\partial y} (y) + y z \times \frac{\partial}{\partial y} (e^{x y z})$ $= {z e}^{x y z} + y z \times e^{x y z} \times x z$ $= e^{x y z} (z + {x y z}^{2})$ $u_{x y z} = e^{x y z} \times \frac{\partial}{\partial z} (z + {x y z}^{2}) + (z + {x y z}^{2}) \times \frac{\partial}{\partial z} (e^{x y z})$ $= e^{x y z} \times (1 + 2 x y z) + (z + {x y z}^{2}) \times e^{x y z} (x y)$ $= u [1 + 2 x y z + x y z + x^{2} y^{2} z^{2}]$ $= u [1 + 3 x y z + x^{2} y^{2} z^{2}] s o o p t i o n a i s c o r r e c t$
	Commented by Necxx last updated on 21/Jan/19

	$e x a c t l y . . . . . T h a n k s s o m u c h$
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