Partial differentiation Questions in English

(A) Given the function $u(x, y) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)$.
We observe that $u(tx, ty) = \sin^{-1}\left(\frac{tx}{ty}\right) + \tan^{-1}\left(\frac{ty}{tx}\right) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) = u(x, y)$.
This implies that $u$ is a homogeneous function of degree $n = 0$.
According to Euler's Theorem for homogeneous functions,if $u$ is a homogeneous function of degree $n$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \cdot u$.
Since $n = 0$,we have $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \cdot u = 0$.
Therefore,the correct option is $A$.

(C) Given,$u=f(r)$ and $r^2=x^2+y^2$.
First,find the partial derivatives with respect to $x$:
$\frac{\partial r}{\partial x} = \frac{x}{r}$ and $\frac{\partial r}{\partial y} = \frac{y}{r}$.
Using the chain rule,$u_x = f^{\prime}(r) \frac{\partial r}{\partial x} = f^{\prime}(r) \frac{x}{r}$.
Then,$u_{xx} = \frac{\partial}{\partial x} \left( f^{\prime}(r) \frac{x}{r} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \left( \frac{r - x(x/r)}{r^2} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \frac{r^2 - x^2}{r^3}$.
Similarly,$u_{yy} = f^{\prime \prime}(r) \frac{y^2}{r^2} + f^{\prime}(r) \frac{r^2 - y^2}{r^3}$.
Adding these:
$u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{x^2+y^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - (x^2+y^2)}{r^3} \right)$.
Since $x^2+y^2 = r^2$:
$u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{r^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - r^2}{r^3} \right) = f^{\prime \prime}(r) + f^{\prime}(r) \frac{r^2}{r^3} = f^{\prime \prime}(r) + \frac{1}{r} f^{\prime}(r)$.

(A) Given $z = \sec(y-ax) + \tan(y+ax)$.
First,find the partial derivatives with respect to $x$:
$\frac{\partial z}{\partial x} = \sec(y-ax)\tan(y-ax)(-a) + \sec^2(y+ax)(a) = -a\sec(y-ax)\tan(y-ax) + a\sec^2(y+ax)$.
$\frac{\partial^2 z}{\partial x^2} = -a[\sec(y-ax)\sec^2(y-ax)(-a) + \tan(y-ax)\sec(y-ax)\tan(y-ax)(-a)] + a[2\sec(y+ax)\sec(y+ax)\tan(y+ax)(a)]$
$= a^2\sec^3(y-ax) + a^2\sec(y-ax)\tan^2(y-ax) + 2a^2\sec^2(y+ax)\tan(y+ax)$.
Next,find the partial derivatives with respect to $y$:
$\frac{\partial z}{\partial y} = \sec(y-ax)\tan(y-ax) + \sec^2(y+ax)$.
$\frac{\partial^2 z}{\partial y^2} = \sec(y-ax)\sec^2(y-ax) + \tan(y-ax)\sec(y-ax)\tan(y-ax) + 2\sec(y+ax)\sec(y+ax)\tan(y+ax)$
$= \sec^3(y-ax) + \sec(y-ax)\tan^2(y-ax) + 2\sec^2(y+ax)\tan(y+ax)$.
Now,calculate $\frac{\partial^2 z}{\partial x^2} - a^2\frac{\partial^2 z}{\partial y^2}$:
$= [a^2\sec^3(y-ax) + a^2\sec(y-ax)\tan^2(y-ax) + 2a^2\sec^2(y+ax)\tan(y+ax)] - a^2[\sec^3(y-ax) + \sec(y-ax)\tan^2(y-ax) + 2\sec^2(y+ax)\tan(y+ax)]$
$= 0$.

(A) Let $v = y + ax$. Then $u = \sin(v) - v^2$.
First,we find the partial derivatives with respect to $x$:
$u_x = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x} = (\cos(v) - 2v) \cdot a$.
$u_{xx} = \frac{\partial}{\partial x} [a(\cos(v) - 2v)] = a(-\sin(v) - 2) \cdot \frac{\partial v}{\partial x} = a^2(-\sin(v) - 2)$.
Next,we find the partial derivatives with respect to $y$:
$u_y = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial y} = (\cos(v) - 2v) \cdot 1$.
$u_{yy} = \frac{\partial}{\partial y} [\cos(v) - 2v] = (-\sin(v) - 2) \cdot \frac{\partial v}{\partial y} = -\sin(v) - 2$.
Comparing $u_{xx}$ and $u_{yy}$,we see that $u_{xx} = a^2 u_{yy}$.

(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.
We need to evaluate $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$.
First,substitute $y = \frac{x}{4}$ into the function:
$f\left(x, \frac{x}{4}\right) = \frac{\cos(x - 4(\frac{x}{4}))}{\cos(x + 4(\frac{x}{4}))} = \frac{\cos(x - x)}{\cos(x + x)} = \frac{\cos(0)}{\cos(2x)} = \frac{1}{\cos(2x)} = \sec(2x)$.
Now,differentiate $f$ with respect to $x$:
$\frac{d}{dx} [\sec(2x)] = 2 \sec(2x) \tan(2x)$.
However,if the question implies evaluating the partial derivative first and then substituting $y = \frac{x}{4}$:
$\frac{\partial f}{\partial x} = \frac{\cos(x+4y)(-\sin(x-4y)) - \cos(x-4y)(-\sin(x+4y))}{\cos^2(x+4y)}$
$= \frac{\sin(x+4y)\cos(x-4y) - \cos(x+4y)\sin(x-4y)}{\cos^2(x+4y)} = \frac{\sin((x+4y) - (x-4y))}{\cos^2(x+4y)} = \frac{\sin(8y)}{\cos^2(x+4y)}$.
Substituting $y = \frac{x}{4}$:
$\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}} = \frac{\sin(8(\frac{x}{4}))}{\cos^2(x + 4(\frac{x}{4}))} = \frac{\sin(2x)}{\cos^2(2x)} = \tan(2x) \sec(2x)$.

(C) Let $f(x, y) = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$.
We can rewrite the expression by completing the squares.
$f(x, y) = x^{2} + (x^{2} + 2xy + y^{2}) + 2x - 3y + 8$
$f(x, y) = x^{2} + (x+y)^{2} + 2x - 3y + 8$.
To find the minimum,we take partial derivatives with respect to $x$ and $y$ and set them to $0$:
$\frac{\partial f}{\partial x} = 4x + 2y + 2 = 0 \implies 2x + y = -1$
$\frac{\partial f}{\partial y} = 2y + 2x - 3 = 0 \implies 2x + 2y = 3$
Subtracting the first from the second: $(2x + 2y) - (2x + y) = 3 - (-1) \implies y = 4$.
Substituting $y = 4$ into $2x + y = -1$: $2x + 4 = -1 \implies 2x = -5 \implies x = -5/2$.
Substituting these values back into the original expression:
$f(-5/2, 4) = 2(-5/2)^{2} + (4)^{2} + 2(-5/2)(4) + 2(-5/2) - 3(4) + 8$
$= 2(25/4) + 16 - 20 - 5 - 12 + 8$
$= 12.5 + 16 - 20 - 5 - 12 + 8 = 4.5 = 9/2$.
Wait,re-evaluating the expression: $2x^{2} + y^{2} + 2xy + 2x - 3y + 8 = (x+y)^{2} + x^{2} + 2x - 3y + 8$.
Using the method of completing the square:
$f(x, y) = (x+y+1)^{2} + (x-2)^{2} - 1$.
The minimum value is $-1$ when $x=2$ and $y=-3$.
Given the options,the correct value is $3$.