If u=e^(x//y)+e^(y//z)+e^(z//x) then show that x*(del u)/(del x)+y*(del u)/(del y del+z,del z)=0 x//y((x)/(y)+xy)+e^(y//2)((y)/(x)+2y)+e^(z//x)(x_(1)z+z)

Question

If  u=e^(x//y)+e^(y//z)+e^(z//x) then show that  x*(del u)/(del x)+y*(del u)/(del y del+z,del z)=0  x//y((x)/(y)+xy)+e^(y//2)((y)/(x)+2y)+e^(z//x)(x_(1)z+z)

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Find Partial Derivative of u with Respect to x: Let's start by finding the partial derivative of u with respect to x. u = e^(x/y) + e^(y/z) + e^(z/x) (∂u/∂x) = (∂/∂x)(e^(x/y)) + (∂/∂x)(e^(y/z)) + (∂/∂x)(e^(z/x)) Since y and z are treated as constants when taking the partial derivative with respect to x, the second and third terms do not depend on x and their derivatives are zero. (∂/∂x)(e^(x/y)) = (1/y)e^(x/y) * (dx/dx) = (1/y)e^(x/y) So, (∂u/∂x) = (1/y)e^(x/y)Find Partial Derivative of u with Respect to y: Now let's find the partial derivative of u with respect to y. u = e^(x/y) + e^(y/z) + e^(z/x) (∂u/∂y) = (∂/∂y)(e^(x/y)) + (∂/∂y)(e^(y/z)) + (∂/∂y)(e^(z/x)) The first term is a function of x/y, so we use the chain rule: (∂/∂y)(e^(x/y)) = -x/y^2 * e^(x/y) The second term is a function of y/z, so its derivative is (1/z)e^(y/z) The third term does not depend on y, so its derivative is zero. So, (∂u/∂y) = -x/y^2 * e^(x/y) + (1/z)e^(y/z)Find Partial Derivative of u with Respect to z: Next, we find the partial derivative of u with respect to z. u = e^(x/y) + e^(y/z) + e^(z/x) (∂u/∂z) = (∂/∂z)(e^(x/y)) + (∂/∂z)(e^(y/z)) + (∂/∂z)(e^(z/x)) The first term does not depend on z, so its derivative is zero. The second term is a function of y/z, so we use the chain rule: (∂/∂z)(e^(y/z)) = -y/z^2 * e^(y/z) The third term is a function of z/x, so its derivative is (1/x)e^(z/x) So, (∂u/∂z) = -y/z^2 * e^(y/z) + (1/x)e^(z/x)Multiply and Sum Partial Derivatives: Now we multiply each partial derivative by its corresponding variable and sum them up. x*(∂u/∂x) + y*(∂u/∂y) + z*(∂u/∂z) = x*(1/y)e^(x/y) + y*(-x/y^2 * e^(x/y) + (1/z)e^(y/z)) + z*(-y/z^2 * e^(y/z) + (1/x)e^(z/x)) Simplify the expression by combining like terms and factoring out common factors. = e^(x/y) + (-x/y)e^(x/y) + (y/z)e^(y/z) - (y/z)e^(y/z) + (z/x)e^(z/x) = e^(x/y) - (x/y)e^(x/y) + (y/z)e^(y/z) - (y/z)e^(y/z) + (z/x)e^(z/x) Notice that the terms e^(x/y) and -(x/y)e^(x/y) cancel each other out, as do the terms (y/z)e^(y/z) and -(y/z)e^(y/z). So we are left with (z/x)e^(z/x)

If $u=e^{\frac{x}{y}}+e^{\frac{y}{z}}+e^{\frac{z}{x}}$ then show that $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=0$ $\newline$ $e^{\frac{x}{y}}\left(\frac{x}{y}+xy\right)+e^{\frac{y}{z}}\left(\frac{y}{x}+2y\right)+e^{\frac{z}{x}}(x_{1}z+z)$

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