Partial differentiation Questions in English

(D) Given: $u = x^2 + y^2$,$x = s + 3t$,$y = 2s - t$.
First,find the first derivatives with respect to $s$:
$\frac{dx}{ds} = 1$ and $\frac{dy}{ds} = 2$.
Next,find the second derivatives with respect to $s$:
$\frac{d^2x}{ds^2} = 0$ and $\frac{d^2y}{ds^2} = 0$.
Now,differentiate $u$ with respect to $s$ using the chain rule:
$\frac{du}{ds} = 2x \frac{dx}{ds} + 2y \frac{dy}{ds}$.
Differentiate again with respect to $s$:
$\frac{d^2u}{ds^2} = 2 \left( \frac{dx}{ds} \right)^2 + 2x \frac{d^2x}{ds^2} + 2 \left( \frac{dy}{ds} \right)^2 + 2y \frac{d^2y}{ds^2}$.
Substitute the values:
$\frac{d^2u}{ds^2} = 2(1)^2 + 2x(0) + 2(2)^2 + 2y(0) = 2(1) + 8 = 10$.

(D) The given function is $z = \frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1/4}}$.
This is a homogeneous function of $x$ and $y$.
Let $f(x, y) = (x^4 + y^4)^{1/3}$ and $g(x, y) = (x^3 + y^3)^{1/4}$.
The degree of $f(x, y)$ is $n_1 = \frac{4}{3}$ and the degree of $g(x, y)$ is $n_2 = \frac{3}{4}$.
Since $z = \frac{f}{g}$,the degree of $z$ is $n = n_1 - n_2 = \frac{4}{3} - \frac{3}{4} = \frac{16 - 9}{12} = \frac{7}{12}$.
By Euler's Theorem on homogeneous functions,if $z$ is a homogeneous function of degree $n$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = nz$.
Substituting $n = \frac{7}{12}$,we get $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{7}{12}z$.

(C) Given $z = \tan^{-1}\left(\frac{x}{y}\right)$.
First,find the partial derivative of $z$ with respect to $x$ $(z_x)$:
$z_x = \frac{\partial z}{\partial x} = \frac{1}{1 + (x/y)^2} \cdot \frac{1}{y} = \frac{1}{(y^2 + x^2)/y^2} \cdot \frac{1}{y} = \frac{y^2}{x^2 + y^2} \cdot \frac{1}{y} = \frac{y}{x^2 + y^2}$.
Next,find the partial derivative of $z$ with respect to $y$ $(z_y)$:
$z_y = \frac{\partial z}{\partial y} = \frac{1}{1 + (x/y)^2} \cdot \left(-\frac{x}{y^2}\right) = \frac{y^2}{x^2 + y^2} \cdot \left(-\frac{x}{y^2}\right) = -\frac{x}{x^2 + y^2}$.
Now,calculate the ratio $z_x : z_y$:
$z_x : z_y = \frac{y}{x^2 + y^2} : -\frac{x}{x^2 + y^2} = y : -x = -y : x$.
Therefore,the correct option is $C$.

(B) Given $u = \frac{x + y}{x - y}$.
First,find the partial derivative of $u$ with respect to $x$ using the quotient rule:
$\frac{\partial u}{\partial x} = \frac{(x - y)(1) - (x + y)(1)}{(x - y)^2} = \frac{x - y - x - y}{(x - y)^2} = \frac{-2y}{(x - y)^2}$.
Next,find the partial derivative of $u$ with respect to $y$ using the quotient rule:
$\frac{\partial u}{\partial y} = \frac{(x - y)(1) - (x + y)(-1)}{(x - y)^2} = \frac{x - y + x + y}{(x - y)^2} = \frac{2x}{(x - y)^2}$.
Now,add the two partial derivatives:
$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{-2y}{(x - y)^2} + \frac{2x}{(x - y)^2} = \frac{2x - 2y}{(x - y)^2} = \frac{2(x - y)}{(x - y)^2} = \frac{2}{x - y}$.

(B) Given $u = \log ({x^2} + {y^2})$.
First,find the partial derivative with respect to $x$:
$\frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}} \cdot 2x = \frac{{2x}}{{{x^2} + {y^2}}}$.
Now,find the second partial derivative with respect to $x$ using the quotient rule:
$\frac{{\partial ^2}u}{\partial {x^2}} = \frac{{({x^2} + {y^2}) \cdot 2 - 2x \cdot 2x}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{x^2} + 2{y^2} - 4{x^2}}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$.
Similarly,find the partial derivative with respect to $y$:
$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}} \cdot 2y = \frac{{2y}}{{{x^2} + {y^2}}}$.
Now,find the second partial derivative with respect to $y$:
$\frac{{\partial ^2}u}{\partial {y^2}} = \frac{{({x^2} + {y^2}) \cdot 2 - 2y \cdot 2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{x^2} + 2{y^2} - 4{y^2}}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$.
Adding these two results:
$\frac{{\partial ^2}u}{\partial {x^2}} + \frac{{\partial ^2}u}{\partial {y^2}} = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}} + \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{y^2} - 2{x^2} + 2{x^2} - 2{y^2}}}{{{{({x^2} + {y^2})}^2}}} = 0$.

(D) Given $u = \sin^{-1}\left(\frac{y}{x}\right)$.
To find $\frac{\partial u}{\partial x}$,we differentiate $u$ with respect to $x$ while treating $y$ as a constant.
Using the chain rule,$\frac{d}{dx}(\sin^{-1}(v)) = \frac{1}{\sqrt{1 - v^2}} \cdot \frac{dv}{dx}$.
Here,$v = \frac{y}{x}$,so $\frac{\partial v}{\partial x} = y \cdot \frac{d}{dx}(x^{-1}) = y \cdot (-x^{-2}) = -\frac{y}{x^2}$.
Substituting these into the derivative formula:
$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - (\frac{y}{x})^2}} \cdot \left(-\frac{y}{x^2}\right)$
$= \frac{1}{\sqrt{\frac{x^2 - y^2}{x^2}}} \cdot \left(-\frac{y}{x^2}\right)$
$= \frac{x}{\sqrt{x^2 - y^2}} \cdot \left(-\frac{y}{x^2}\right)$
$= -\frac{y}{x\sqrt{x^2 - y^2}}$.

(A) Given $u = \tan^{-1}(\frac{y}{x})$.
We can rewrite this as $\tan u = \frac{y}{x}$.
Let $f(u) = \tan u = \frac{y}{x}$.
Here,$f(u)$ is a homogeneous function of $x$ and $y$ with degree $n = 0$ because $\frac{ty}{tx} = (\frac{y}{x})^0 \cdot \frac{y}{x}$.
According to Euler's Theorem for homogeneous functions,if $f(u)$ is a homogeneous function of degree $n$,then $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(u)$.
Here,$f(u) = \tan u$ and $n = 0$.
So,$x \frac{\partial (\tan u)}{\partial x} + y \frac{\partial (\tan u)}{\partial y} = 0 \cdot \tan u = 0$.
Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 0$.
Dividing by $\sec^2 u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$.

(A) Given $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$.
This implies $\tan u = \frac{x^3 + y^3}{x - y}$.
Let $f(x, y) = \tan u = \frac{x^3 + y^3}{x - y}$.
Here,$f(x, y)$ is a homogeneous function of degree $n = 3 - 1 = 2$.
According to Euler's theorem for homogeneous functions,$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f$.
Substituting $f = \tan u$ and $n = 2$,we get $x\frac{\partial}{\partial x}(\tan u) + y\frac{\partial}{\partial y}(\tan u) = 2 \tan u$.
Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 2 \tan u$.
Dividing by $\sec^2 u$,we get $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2 \frac{\tan u}{\sec^2 u}$.
Since $\frac{\tan u}{\sec^2 u} = \frac{\sin u / \cos u}{1 / \cos^2 u} = \sin u \cos u$,we have $2 \sin u \cos u = \sin 2u$.

(C) According to Euler's theorem for homogeneous functions, if $F(u)$ is a homogeneous function of degree $n$ in $x, y, z$, then:
$x \frac{\partial}{\partial x}(F(u)) + y \frac{\partial}{\partial y}(F(u)) + z \frac{\partial}{\partial z}(F(u)) = n F(u)$
Using the chain rule for partial differentiation, we have $\frac{\partial}{\partial x}(F(u)) = F'(u) \frac{\partial u}{\partial x}$, $\frac{\partial}{\partial y}(F(u)) = F'(u) \frac{\partial u}{\partial y}$, and $\frac{\partial}{\partial z}(F(u)) = F'(u) \frac{\partial u}{\partial z}$.
Substituting these into the equation:
$x F'(u) \frac{\partial u}{\partial x} + y F'(u) \frac{\partial u}{\partial y} + z F'(u) \frac{\partial u}{\partial z} = n F(u)$
Factoring out $F'(u)$:
$F'(u) \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} \right) = n F(u)$
Dividing by $F'(u)$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = \frac{n F(u)}{F'(u)}$
Thus, the correct option is $C$.

(D) Given $u = \log (x^3 + y^3 + z^3 - 3xyz)$.
We know that $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.
Therefore,$u = \log(x + y + z) + \log(x^2 + y^2 + z^2 - xy - yz - zx)$.
Calculating the partial derivatives:
$\frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz}$
$\frac{\partial u}{\partial y} = \frac{3y^2 - 3zx}{x^3 + y^3 + z^3 - 3xyz}$
$\frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}$
Summing these derivatives:
$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}$
$= \frac{3}{x + y + z}$
Multiplying by $(x + y + z)$:
$(x + y + z) \left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right) = 3$.

(B) Let $u = \sin z = \frac{x+y}{\sqrt{x} + \sqrt{y}}$.
We check if $u$ is a homogeneous function of $x$ and $y$.
$u(tx, ty) = \frac{tx+ty}{\sqrt{tx} + \sqrt{ty}} = \frac{t(x+y)}{\sqrt{t}(\sqrt{x} + \sqrt{y})} = t^{1/2} u(x, y)$.
Thus,$u$ is a homogeneous function of degree $n = 1/2$.
By Euler's Theorem on homogeneous functions,$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = n u$.
Substituting $u = \sin z$ and $n = 1/2$,we get:
$x\frac{\partial}{\partial x}(\sin z) + y\frac{\partial}{\partial y}(\sin z) = \frac{1}{2} \sin z$.
Applying the chain rule,$x(\cos z \frac{\partial z}{\partial x}) + y(\cos z \frac{\partial z}{\partial y}) = \frac{1}{2} \sin z$.
Dividing both sides by $\cos z$,we obtain:
$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{1}{2} \frac{\sin z}{\cos z} = \frac{1}{2} \tan z$.

(A) Given $u = \log_e(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right)$.
First,find the partial derivatives with respect to $x$:
$\frac{\partial u}{\partial x} = \frac{2x}{x^2 + y^2} + \frac{1}{1 + (y/x)^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{2x}{x^2 + y^2} - \frac{y}{x^2 + y^2} = \frac{2x - y}{x^2 + y^2}$.
Now,find the second partial derivative with respect to $x$:
$\frac{\partial^2 u}{\partial x^2} = \frac{(x^2 + y^2)(2) - (2x - y)(2x)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4x^2 + 2xy}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2 + 2xy}{(x^2 + y^2)^2}$.
Next,find the partial derivatives with respect to $y$:
$\frac{\partial u}{\partial y} = \frac{2y}{x^2 + y^2} + \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}\right) = \frac{2y}{x^2 + y^2} + \frac{x}{x^2 + y^2} = \frac{2y + x}{x^2 + y^2}$.
Now,find the second partial derivative with respect to $y$:
$\frac{\partial^2 u}{\partial y^2} = \frac{(x^2 + y^2)(2) - (2y + x)(2y)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4y^2 - 2xy}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2 - 2xy}{(x^2 + y^2)^2}$.
Adding the two second partial derivatives:
$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2y^2 - 2x^2 + 2xy + 2x^2 - 2y^2 - 2xy}{(x^2 + y^2)^2} = \frac{0}{(x^2 + y^2)^2} = 0$.

(B) Given the equation: ${x^x}{y^y}{z^z} = c$
Taking the natural logarithm on both sides: $\log({x^x}{y^y}{z^z}) = \log c$
Using logarithmic properties: $x\log x + y\log y + z\log z = \log c$ .....$(i)$
Here,$x$ and $y$ are independent variables,and $z$ is a function of $x$ and $y$.
Differentiating equation $(i)$ partially with respect to $x$:
$\frac{\partial}{\partial x}(x\log x) + \frac{\partial}{\partial x}(y\log y) + \frac{\partial}{\partial x}(z\log z) = \frac{\partial}{\partial x}(\log c)$
$(1 \cdot \log x + x \cdot \frac{1}{x}) + 0 + (1 \cdot \log z + z \cdot \frac{1}{z}) \frac{\partial z}{\partial x} = 0$
$(1 + \log x) + (1 + \log z) \frac{\partial z}{\partial x} = 0$
Rearranging the terms to solve for $\frac{\partial z}{\partial x}$:
$(1 + \log z) \frac{\partial z}{\partial x} = -(1 + \log x)$
$\frac{\partial z}{\partial x} = -\frac{1 + \log x}{1 + \log z}$
Thus,the correct option is $B$.

(C) Given the function $u(x, y) = x{y^2}{\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$.
To check for homogeneity,we replace $x$ with $tx$ and $y$ with $ty$:
$u(tx, ty) = (tx)(ty)^2 \tan^{-1}\left(\frac{ty}{tx}\right)$
$u(tx, ty) = t^3 x y^2 \tan^{-1}\left(\frac{y}{x}\right)$
$u(tx, ty) = t^3 u(x, y)$.
Since the function is homogeneous of degree $n = 3$,by Euler's Theorem on homogeneous functions,we have:
$x{u_x} + y{u_y} = n u$
$x{u_x} + y{u_y} = 3u$.

(D) Let $f(x, y) = z^2 = \frac{x^{1/2} + y^{1/2}}{x^{1/3} + y^{1/3}}$.
Check for homogeneity: $f(tx, ty) = \frac{(tx)^{1/2} + (ty)^{1/2}}{(tx)^{1/3} + (ty)^{1/3}} = \frac{t^{1/2}(x^{1/2} + y^{1/2})}{t^{1/3}(x^{1/3} + y^{1/3})} = t^{1/2 - 1/3} f(x, y) = t^{1/6} f(x, y)$.
Since $z^2$ is a homogeneous function of degree $n = \frac{1}{6}$,by Euler's Theorem on homogeneous functions:
$x \frac{\partial (z^2)}{\partial x} + y \frac{\partial (z^2)}{\partial y} = n z^2 = \frac{1}{6} z^2$.
Using the chain rule,$\frac{\partial (z^2)}{\partial x} = 2z \frac{\partial z}{\partial x}$ and $\frac{\partial (z^2)}{\partial y} = 2z \frac{\partial z}{\partial y}$.
Substituting these into the equation:
$x(2z \frac{\partial z}{\partial x}) + y(2z \frac{\partial z}{\partial y}) = \frac{1}{6} z^2$.
Dividing both sides by $2z$ (assuming $z \neq 0$):
$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{1}{12} z$.

(B) Given $u = \tan^{-1}(x + y)$,which implies $\tan u = x + y$.
Let $f(x, y) = \tan u = x + y$.
Since $f(x, y)$ is a homogeneous function of degree $n = 1$ in $x$ and $y$,by Euler's Theorem for homogeneous functions,we have:
$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n \cdot f(x, y)$
$x\frac{\partial}{\partial x}(\tan u) + y\frac{\partial}{\partial y}(\tan u) = 1 \cdot \tan u$
$x(\sec^2 u)\frac{\partial u}{\partial x} + y(\sec^2 u)\frac{\partial u}{\partial y} = \tan u$
Dividing both sides by $\sec^2 u$,we get:
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{\tan u}{\sec^2 u}$
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} \cdot \cos^2 u = \sin u \cos u$
Using the identity $\sin 2u = 2 \sin u \cos u$,we have $\sin u \cos u = \frac{1}{2} \sin 2u$.
Therefore,$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{1}{2} \sin 2u$.

(B) Given $u = (x^2 + y^2 + z^2)^{3/2}$.
First,find the partial derivative with respect to $x$:
$\frac{\partial u}{\partial x} = \frac{3}{2}(x^2 + y^2 + z^2)^{1/2} \cdot (2x) = 3x(x^2 + y^2 + z^2)^{1/2}$.
Squaring this,we get:
$\left( \frac{\partial u}{\partial x} \right)^2 = 9x^2(x^2 + y^2 + z^2)$.
Similarly,by symmetry:
$\left( \frac{\partial u}{\partial y} \right)^2 = 9y^2(x^2 + y^2 + z^2)$ and $\left( \frac{\partial u}{\partial z} \right)^2 = 9z^2(x^2 + y^2 + z^2)$.
Adding these three expressions:
$\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial u}{\partial z} \right)^2 = 9(x^2 + y^2 + z^2)(x^2 + y^2 + z^2 + z^2) = 9(x^2 + y^2 + z^2)^2$.
Since $u = (x^2 + y^2 + z^2)^{3/2}$,we have $u^{2/3} = (x^2 + y^2 + z^2)$.
Therefore,$(x^2 + y^2 + z^2)^2 = (u^{2/3})^2 = u^{4/3}$.
Thus,the sum is $9u^{4/3}$.

(B) Given $u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 \tan^{-1}(\frac{x}{y})$.
Using the identity $\tan^{-1}(\frac{x}{y}) = \frac{\pi}{2} - \tan^{-1}(\frac{y}{x})$ for $y > 0$,we have:
$u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 (\frac{\pi}{2} - \tan^{-1}(\frac{y}{x}))$
$u = (x^2 + y^2) \tan^{-1}(\frac{y}{x}) - \frac{\pi}{2} y^2$
Now,differentiate $u$ partially with respect to $y$:
$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} + 2y \tan^{-1}(\frac{y}{x}) - \pi y$
$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x} + 2y \tan^{-1}(\frac{y}{x}) - \pi y$
$\frac{\partial u}{\partial y} = x + 2y \tan^{-1}(\frac{y}{x}) - \pi y$
Next,differentiate $\frac{\partial u}{\partial y}$ partially with respect to $x$:
$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{1}{1 + (y/x)^2} \cdot (-\frac{y}{x^2}) - 0$
$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{x^2}{x^2 + y^2} \cdot (-\frac{y}{x^2})$
$\frac{\partial^2 u}{\partial x \partial y} = 1 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}$.

(A) Given $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$.
Differentiating with respect to $x$:
$2u \frac{\partial u}{\partial x} = 2(x - a) \implies u \frac{\partial u}{\partial x} = x - a$.
Differentiating again with respect to $x$:
$u \frac{\partial^2 u}{\partial x^2} + \left(\frac{\partial u}{\partial x}\right)^2 = 1$.
Substituting $\frac{\partial u}{\partial x} = \frac{x - a}{u}$:
$u \frac{\partial^2 u}{\partial x^2} = 1 - \left(\frac{x - a}{u}\right)^2 = 1 - \frac{(x - a)^2}{u^2} = \frac{u^2 - (x - a)^2}{u^2}$.
Thus,$\frac{\partial^2 u}{\partial x^2} = \frac{1}{u} - \frac{(x - a)^2}{u^3}$.
Similarly,$\frac{\partial^2 u}{\partial y^2} = \frac{1}{u} - \frac{(y - b)^2}{u^3}$ and $\frac{\partial^2 u}{\partial z^2} = \frac{1}{u} - \frac{(z - c)^2}{u^3}$.
Summing these:
$\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{1}{u^3} [(x - a)^2 + (y - b)^2 + (z - c)^2]$.
Since $(x - a)^2 + (y - b)^2 + (z - c)^2 = u^2$:
$\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{u^2}{u^3} = \frac{3}{u} - \frac{1}{u} = \frac{2}{u}$.

(B) The given function is $z = f\left( \frac{x}{y} \right)$.
This implies that $z$ is a homogeneous function of $x$ and $y$ with degree $n = 0$.
According to Euler's theorem for homogeneous functions,if $z$ is a homogeneous function of degree $n$,then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = nz$.
Since $n = 0$,we have $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0 \cdot z = 0$.
Therefore,$x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

(B) Given $u = e^{-x^2 - y^2}$.
First,find the partial derivative of $u$ with respect to $x$:
$u_x = \frac{\partial u}{\partial x} = e^{-x^2 - y^2} \cdot (-2x) = -2x u$.
Next,find the partial derivative of $u$ with respect to $y$:
$u_y = \frac{\partial u}{\partial y} = e^{-x^2 - y^2} \cdot (-2y) = -2y u$.
Now,consider the expression $y u_x$:
$y u_x = y(-2x u) = -2xy u$.
Consider the expression $x u_y$:
$x u_y = x(-2y u) = -2xy u$.
Comparing both,we get $y u_x = x u_y$.

(C) Given $u(x,y) = y \log x + x \log y$.
First,find the partial derivatives:
${u_x} = \frac{\partial u}{\partial x} = \frac{y}{x} + \log y$
${u_y} = \frac{\partial u}{\partial y} = \log x + \frac{x}{y}$
Now,substitute these into the expression ${u_x}{u_y} - {u_x} \log x - {u_y} \log y + \log x \log y$:
$= (\frac{y}{x} + \log y)(\log x + \frac{x}{y}) - (\frac{y}{x} + \log y)\log x - (\log x + \frac{x}{y})\log y + \log x \log y$
Expand the terms:
$= (\frac{y}{x} \log x + 1 + \log y \log x + \frac{x}{y} \log y) - (\frac{y}{x} \log x + \log y \log x) - (\log x \log y + \frac{x}{y} \log y) + \log x \log y$
Cancel the terms:
$= \frac{y}{x} \log x + 1 + \log y \log x + \frac{x}{y} \log y - \frac{y}{x} \log x - \log y \log x - \log x \log y - \frac{x}{y} \log y + \log x \log y$
$= 1$.

(C) Given $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$.
Let $f(u) = \sin u = \sqrt{\frac{x^2 + y^2}{x + y}}$.
Let $z = f(u) = \sqrt{\frac{x^2 + y^2}{x + y}}$.
We check the degree of homogeneity of $z$. Replacing $x$ with $tx$ and $y$ with $ty$:
$z(tx, ty) = \sqrt{\frac{(tx)^2 + (ty)^2}{tx + ty}} = \sqrt{\frac{t^2(x^2 + y^2)}{t(x + y)}} = t^{1/2} \sqrt{\frac{x^2 + y^2}{x + y}} = t^{1/2} z$.
Thus,$z$ is a homogeneous function of degree $n = 1/2$.
By Euler's theorem for homogeneous functions,$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$.
Since $z = \sin u$,we have $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = \frac{1}{2} \sin u$.
$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \frac{1}{2} \sin u$.
Dividing by $\cos u$,we get $x u_x + y u_y = \frac{1}{2} \frac{\sin u}{\cos u} = \frac{1}{2} \tan u$.

(B) Given $z = y + f(v)$ and $v = \frac{x}{y}$.
Substituting $v$,we have $z = y + f\left(\frac{x}{y}\right)$.
Now,differentiate $z$ partially with respect to $x$:
$\frac{\partial z}{\partial x} = f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}$.
Next,differentiate $z$ partially with respect to $y$ using the chain rule:
$\frac{\partial z}{\partial y} = 1 + f'\left(\frac{x}{y}\right) \cdot \left(-\frac{x}{y^2}\right) = 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.
Now,calculate the expression $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$:
$v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \left(\frac{x}{y}\right) \left[f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}\right] + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.
$= \frac{x}{y^2} f'\left(\frac{x}{y}\right) + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.
$= 1$.

(C) Given the function $f(x, y) = 2(x-y)^2 - x^4 - y^4$.
First,find the partial derivatives with respect to $x$ and $y$:
$f_x = 4(x-y) - 4x^3$
$f_y = -4(x-y) - 4y^3$
Now,find the second-order partial derivatives:
$f_{xx} = \frac{\partial}{\partial x}(4x - 4y - 4x^3) = 4 - 12x^2$
$f_{yy} = \frac{\partial}{\partial y}(-4x + 4y - 4y^3) = 4 - 12y^2$
$f_{xy} = \frac{\partial}{\partial y}(4x - 4y - 4x^3) = -4$
Evaluating these at the point $(0,0)$:
$(f_{xx})_{(0,0)} = 4 - 12(0)^2 = 4$
$(f_{yy})_{(0,0)} = 4 - 12(0)^2 = 4$
$(f_{xy})_{(0,0)} = -4$
Finally,calculate the value of the expression $(f_{xx}f_{yy} - f_{xy}^2)$ at $(0,0)$:
$(f_{xx}f_{yy} - f_{xy}^2)_{(0,0)} = (4)(4) - (-4)^2 = 16 - 16 = 0$.

(D) Given,$u=\log \left(x^3+y^3+z^3-3 x y z\right)$.
We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-x y-y z-z x)$.
So,$u = \log(x+y+z) + \log(x^2+y^2+z^2-x y-y z-z x)$.
Now,calculate the partial derivatives:
$u_x = \frac{\partial u}{\partial x} = \frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$
$u_y = \frac{\partial u}{\partial y} = \frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$
$u_z = \frac{\partial u}{\partial z} = \frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$
Adding these,we get:
$u_x+u_y+u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$
$u_x+u_y+u_z = \frac{3}{x+y+z}$
Therefore,$(x+y+z)(u_x+u_y+u_z) = 3$.

(C) Given that,$u=e^{x^2-y^2}$.
First,find the partial derivative of $u$ with respect to $x$:
$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{x^2-y^2}(2x)$.
Next,find the partial derivative of $u$ with respect to $y$:
$u_y = \frac{\partial}{\partial y}(e^{x^2-y^2}) = e^{x^2-y^2}(-2y)$.
Now,multiply $u_x$ by $y$:
$y u_x = y \cdot e^{x^2-y^2}(2x) = 2xy e^{x^2-y^2}$.
Multiply $u_y$ by $x$:
$x u_y = x \cdot e^{x^2-y^2}(-2y) = -2xy e^{x^2-y^2}$.
Adding these two expressions:
$y u_x + x u_y = 2xy e^{x^2-y^2} - 2xy e^{x^2-y^2} = 0$.
Thus,the correct option is $C$.

(A) Given,$z=\tan (y+a x)+\sqrt{y-a x}$
First,find the partial derivatives with respect to $x$:
$z_x = \frac{\partial z}{\partial x} = a \sec^2(y+ax) - \frac{a}{2\sqrt{y-ax}}$
$z_{xx} = \frac{\partial^2 z}{\partial x^2} = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}$
Next,find the partial derivatives with respect to $y$:
$z_y = \frac{\partial z}{\partial y} = \sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}$
$z_{yy} = \frac{\partial^2 z}{\partial y^2} = 2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}$
Now,calculate $z_{xx} - a^2 z_{yy}$:
$z_{xx} - a^2 z_{yy} = [2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}] - a^2 [2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}]$
$z_{xx} - a^2 z_{yy} = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}} - 2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{a^2}{4(y-ax)^{3/2}} = 0$

(C) Given,$u(x, y)=y \log x+x \log y$.
First,we find the partial derivatives $u_x$ and $u_y$:
$u_x = \frac{\partial}{\partial x}(y \log x + x \log y) = \frac{y}{x} + \log y$.
$u_y = \frac{\partial}{\partial y}(y \log x + x \log y) = \log x + \frac{x}{y}$.
Now,consider the expression $E = u_x u_y - u_x \log x - u_y \log y + \log x \log y$.
We can factor this expression as:
$E = u_x(u_y - \log x) - \log y(u_y - \log x)$.
$E = (u_x - \log y)(u_y - \log x)$.
Substitute the values of $u_x$ and $u_y$:
$u_x - \log y = (\frac{y}{x} + \log y) - \log y = \frac{y}{x}$.
$u_y - \log x = (\log x + \frac{x}{y}) - \log x = \frac{x}{y}$.
Therefore,$E = (\frac{y}{x}) \times (\frac{x}{y}) = 1$.

(D) Given $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$.
Let $v=\sin u=\frac{x^4+y^4}{x+y}$.
Here,$v$ is a homogeneous function of $x$ and $y$ with degree $n = 4 - 1 = 3$.
According to Euler's theorem for homogeneous functions,$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v$.
Substituting $v = \sin u$ and $n = 3$,we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 3 \sin u$.
Applying the chain rule,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = 3 \sin u$.
Dividing both sides by $\cos u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \frac{\sin u}{\cos u}$.
Therefore,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \tan u$.

(C) The function $u(x, y) = x y^2 \tan^{-1}\left(\frac{y}{x}\right)$ is a homogeneous function of degree $n = 3$ because $u(tx, ty) = (tx)(ty)^2 \tan^{-1}\left(\frac{ty}{tx}\right) = t^3 x y^2 \tan^{-1}\left(\frac{y}{x}\right) = t^3 u(x, y)$.
By Euler's Theorem on homogeneous functions,if $u$ is a homogeneous function of degree $n$ in $x$ and $y$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$.
Here,$n = 3$,so $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 u$.

(B) Given $z = \log(\tan x + \tan y)$.
First,we find the partial derivatives of $z$ with respect to $x$ and $y$:
$\frac{\partial z}{\partial x} = \frac{\sec^2 x}{\tan x + \tan y}$
$\frac{\partial z}{\partial y} = \frac{\sec^2 y}{\tan x + \tan y}$
Now,substitute these into the expression $(\sin 2x) \frac{\partial z}{\partial x} + (\sin 2y) \frac{\partial z}{\partial y}$:
$= \sin 2x \left( \frac{\sec^2 x}{\tan x + \tan y} \right) + \sin 2y \left( \frac{\sec^2 y}{\tan x + \tan y} \right)$
$= \frac{(2 \sin x \cos x) \cdot \frac{1}{\cos^2 x} + (2 \sin y \cos y) \cdot \frac{1}{\cos^2 y}}{\tan x + \tan y}$
$= \frac{2 \tan x + 2 \tan y}{\tan x + \tan y}$
$= \frac{2(\tan x + \tan y)}{\tan x + \tan y} = 2$.

(A) Given $u = \sin(y + ax) - (y + ax)^2$ ... $(i)$
Differentiating partially with respect to $x$:
$u_x = \cos(y + ax) \cdot a - 2(y + ax) \cdot a$
$u_x = a[\cos(y + ax) - 2(y + ax)]$
Differentiating again with respect to $x$:
$u_{xx} = a[-\sin(y + ax) \cdot a - 2 \cdot a]$
$u_{xx} = -a^2[\sin(y + ax) + 2]$ ... (ii)
Differentiating $(i)$ partially with respect to $y$:
$u_y = \cos(y + ax) - 2(y + ax)$
Differentiating again with respect to $y$:
$u_{yy} = -\sin(y + ax) - 2$
$u_{yy} = -[\sin(y + ax) + 2]$ ... (iii)
From (ii) and (iii),we can see that:
$u_{xx} = a^2 \cdot [-\sin(y + ax) - 2]$
$u_{xx} = a^2 \cdot u_{yy}$

(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.
Substitute $y = \frac{x}{2}$ into the function:
$f(x, \frac{x}{2}) = \frac{\cos(x - 4(\frac{x}{2}))}{\cos(x + 4(\frac{x}{2}))} = \frac{\cos(x - 2x)}{\cos(x + 2x)} = \frac{\cos(-x)}{\cos(3x)} = \frac{\cos(x)}{\cos(3x)}$.
Now,calculate the partial derivative with respect to $x$:
$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\cos x}{\cos 3x} \right)$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$\frac{\partial f}{\partial x} = \frac{(-\sin x)(\cos 3x) - (\cos x)(-3 \sin 3x)}{\cos^2 3x} = \frac{3 \cos x \sin 3x - \sin x \cos 3x}{\cos^2 3x}$.
At $y = \frac{x}{2}$,the expression is a function of $x$ only,and its derivative with respect to $x$ is evaluated as shown above.

(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$.
This implies $\sin u = \frac{x^2+y^2}{x+y}$.
Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.
Here,$f(x, y)$ is a homogeneous function of degree $n=1$ because $f(tx, ty) = \frac{(tx)^2+(ty)^2}{tx+ty} = t \cdot \frac{x^2+y^2}{x+y} = t^1 f(x, y)$.
According to Euler's theorem for homogeneous functions,$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f$.
Substituting $f = \sin u$ and $n = 1$,we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 1 \cdot \sin u$.
Applying the chain rule,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \sin u$.
Dividing both sides by $\cos u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} = \tan u$.

(B) Given $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$.
First,calculate $\frac{\partial z}{\partial x}$:
$\frac{\partial z}{\partial x} = \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \frac{1}{y} - \sin \left( 1 + \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right) \right] - \frac{y}{x^2} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$
$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$
$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z \quad \dots (i)$
Next,calculate $\frac{\partial z}{\partial y}$:
$\frac{\partial z}{\partial y} = \frac{1}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] + \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$
$y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right)$
$y \frac{\partial z}{\partial y} = z - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right) \quad \dots (ii)$
Comparing $(i)$ and $(ii)$,we observe that $x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

(B) Let $f(x, y) = \sec z = \frac{x^4+y^4-8x^2y^2}{x^2+y^2}$.
Since $f(x, y)$ is a homogeneous function of degree $n = 4 - 2 = 2$,by Euler's Theorem for homogeneous functions,we have $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y) = 2 \sec z$.
Now,differentiating $f = \sec z$ with respect to $x$ and $y$,we get $\frac{\partial f}{\partial x} = \sec z \tan z \frac{\partial z}{\partial x}$ and $\frac{\partial f}{\partial y} = \sec z \tan z \frac{\partial z}{\partial y}$.
Substituting these into Euler's equation:
$x (\sec z \tan z \frac{\partial z}{\partial x}) + y (\sec z \tan z \frac{\partial z}{\partial y}) = 2 \sec z$.
Dividing both sides by $\sec z \tan z$,we get:
$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{2 \sec z}{\sec z \tan z} = 2 \cot z$.