Differentiation of implicit function Questions in English

(B) Given the equation of the hyperbola: $xy = a$.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y = 0$
Solving for the derivative:
$\frac{dy}{dx} = -\frac{y}{x}$
The slope $m$ of the tangent at the point $(a, 1)$ is obtained by substituting $x = a$ and $y = 1$:
$m = \left( \frac{dy}{dx} \right)_{(a, 1)} = -\frac{1}{a}$.

(B) Given $y = \sqrt{(1 - x)(1 + x)}$.
Squaring both sides,we get $y^2 = (1 - x)(1 + x) = 1 - x^2$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(1 - x^2)$.
Using the chain rule,$2y \frac{dy}{dx} = -2x$.
Dividing by $2$,we get $y \frac{dy}{dx} = -x$.
Multiply both sides by $y$,we get $y^2 \frac{dy}{dx} = -xy$.
Since $y^2 = 1 - x^2$,substituting this gives $(1 - x^2) \frac{dy}{dx} = -xy$.
Rearranging the terms,we get $(1 - x^2) \frac{dy}{dx} + xy = 0$.

(B) Given the equation: ${x^{2/3}} + {y^{2/3}} = {a^{2/3}}$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}({x^{2/3}}) + \frac{d}{dx}({y^{2/3}}) = \frac{d}{dx}({a^{2/3}})$
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule for $y$:
$\frac{2}{3}{x^{-1/3}} + \frac{2}{3}{y^{-1/3}}\frac{dy}{dx} = 0$
Dividing by $\frac{2}{3}$:
${x^{-1/3}} + {y^{-1/3}}\frac{dy}{dx} = 0$
Rearranging to solve for $\frac{dy}{dx}$:
${y^{-1/3}}\frac{dy}{dx} = -{x^{-1/3}}$
$\frac{dy}{dx} = -\frac{{x^{-1/3}}}{{y^{-1/3}}}$
$\frac{dy}{dx} = -{\left( \frac{x}{y} \right)^{-1/3}}$
$\frac{dy}{dx} = -{\left( \frac{y}{x} \right)^{1/3}}$
Thus,the correct option is $B$.

(A) Given equation: $y\sqrt{x^2 + 1} = \log \{\sqrt{x^2 + 1} - x\}$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} \cdot \sqrt{x^2 + 1} + y \cdot \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{2x}{2\sqrt{x^2 + 1}} - 1 \right)$
Simplify the right side:
$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)$
$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{-( \sqrt{x^2 + 1} - x )}{(\sqrt{x^2 + 1} - x) \sqrt{x^2 + 1}}$
$\frac{dy}{dx} \sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = -\frac{1}{\sqrt{x^2 + 1}}$
Multiply the entire equation by $\sqrt{x^2 + 1}$:
$(x^2 + 1)\frac{dy}{dx} + xy = -1$
Rearranging the terms:
$(x^2 + 1)\frac{dy}{dx} + xy + 1 = 0$

(C) Given the curve equation: $\sqrt{x} + \sqrt{y} = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(1)$
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$
Now,substitute the point $\left( \frac{1}{4}, \frac{1}{4} \right)$ into the derivative:
$\left( \frac{dy}{dx} \right)_{\left( \frac{1}{4}, \frac{1}{4} \right)} = -\frac{\sqrt{1/4}}{\sqrt{1/4}} = -\frac{1/2}{1/2} = -1$.

(B) Given the equation $y = \sqrt{\sin x + y}.$
Squaring both sides,we get $y^2 = \sin x + y.$
Differentiating both sides with respect to $x$,we have:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$
$2y \cdot \frac{dy}{dx} = \cos x + \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2y \cdot \frac{dy}{dx} - \frac{dy}{dx} = \cos x$
$\frac{dy}{dx}(2y - 1) = \cos x$
Therefore,$\frac{dy}{dx} = \frac{\cos x}{2y - 1}.$

(C) Given equation: $x = y\sqrt{1 - y^2}$
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$1 = \frac{dy}{dx} \cdot \sqrt{1 - y^2} + y \cdot \frac{1}{2\sqrt{1 - y^2}} \cdot (-2y) \cdot \frac{dy}{dx}$
Simplify the expression:
$1 = \frac{dy}{dx} \left( \sqrt{1 - y^2} - \frac{y^2}{\sqrt{1 - y^2}} \right)$
Find a common denominator inside the bracket:
$1 = \frac{dy}{dx} \left( \frac{1 - y^2 - y^2}{\sqrt{1 - y^2}} \right)$
$1 = \frac{dy}{dx} \left( \frac{1 - 2y^2}{\sqrt{1 - y^2}} \right)$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - 2y^2}$

(A) Given $x = \exp \left\{ {{{\tan }^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} \right)} \right\}$.
Taking natural logarithm on both sides,we get $\log x = {\tan ^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} \right)$.
This implies $\frac{{y - {x^2}}}{{{x^2}}} = \tan (\log x)$.
Rearranging for $y$,we get $y = {x^2}\tan (\log x) + {x^2}$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^2}\tan (\log x)) + \frac{d}{{dx}}({x^2})$.
$\frac{{dy}}{{dx}} = [2x \cdot \tan (\log x) + {x^2} \cdot {\sec ^2}(\log x) \cdot \frac{1}{x}] + 2x$.
$\frac{{dy}}{{dx}} = 2x\tan (\log x) + x{\sec ^2}(\log x) + 2x$.
Factoring out $2x$ from the first and last terms,we get $\frac{{dy}}{{dx}} = 2x[1 + \tan (\log x)] + x{\sec ^2}(\log x)$.

(C) Given the equation: $\sin y + e^{-x \cos y} = e$.
Differentiating both sides with respect to $x$ using the chain rule:
$\cos y \frac{dy}{dx} + e^{-x \cos y} \cdot \frac{d}{dx}(-x \cos y) = 0$
$\cos y \frac{dy}{dx} + e^{-x \cos y} \cdot [(-1) \cos y + (-x)(-\sin y) \frac{dy}{dx}] = 0$
$\cos y \frac{dy}{dx} + e^{-x \cos y} [-\cos y + x \sin y \frac{dy}{dx}] = 0$
$\frac{dy}{dx} (\cos y + x \sin y e^{-x \cos y}) = \cos y e^{-x \cos y}$
$\frac{dy}{dx} = \frac{\cos y e^{-x \cos y}}{\cos y + x \sin y e^{-x \cos y}}$
Now,substitute $(x, y) = (1, \pi)$:
$\cos \pi = -1$ and $\sin \pi = 0$.
$\frac{dy}{dx} \Big|_{(1, \pi)} = \frac{(-1) e^{-1(-1)}}{-1 + 1(0) e^{-1(-1)}} = \frac{-e}{-1} = e$.

(B) Given the equation: ${x^m}{y^n} = {(x + y)^{m + n}}$
Taking the natural logarithm on both sides:
$m \ln x + n \ln y = (m + n) \ln(x + y)$
Differentiating both sides with respect to $x$:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = (m + n) \frac{1}{x + y} \left( 1 + \frac{dy}{dx} \right)$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{n}{y} - \frac{m + n}{x + y} \right) = \frac{m + n}{x + y} - \frac{m}{x}$
$\frac{dy}{dx} \left( \frac{nx + ny - my - ny}{y(x + y)} \right) = \frac{mx + nx - mx - my}{x(x + y)}$
$\frac{dy}{dx} \left( \frac{nx - my}{y(x + y)} \right) = \frac{nx - my}{x(x + y)}$
$\frac{dy}{dx} = \frac{y}{x}$
At $x = 1$ and $y = 2$,we have:
${\left. {\frac{dy}{dx}} \right|_{x = 1, y = 2}} = \frac{2}{1} = 2$.

(A) Given the equation: $\cos (x + y) = y\sin x$
Differentiating both sides with respect to $x$ using the chain rule and product rule:
$-\sin (x + y) \cdot \frac{d}{dx}(x + y) = y \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{dy}{dx}$
$-\sin (x + y) \left( 1 + \frac{dy}{dx} \right) = y\cos x + \sin x \frac{dy}{dx}$
Expanding the left side:
$-\sin (x + y) - \sin (x + y) \frac{dy}{dx} = y\cos x + \sin x \frac{dy}{dx}$
Grouping the $\frac{dy}{dx}$ terms on one side:
$-\sin (x + y) - y\cos x = \sin x \frac{dy}{dx} + \sin (x + y) \frac{dy}{dx}$
$-(\sin (x + y) + y\cos x) = \frac{dy}{dx} (\sin x + \sin (x + y))$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = - \frac{\sin (x + y) + y\cos x}{\sin x + \sin (x + y)}$
Thus,the correct option is $A$.

(D) Given equation: $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$
Rearranging the terms: $x\sqrt{1 + y} = -y\sqrt{1 + x}$
Squaring both sides: $x^2(1 + y) = y^2(1 + x)$
$x^2 + x^2y = y^2 + y^2x$
$x^2 - y^2 + x^2y - y^2x = 0$
$(x - y)(x + y) + xy(x - y) = 0$
$(x - y)(x + y + xy) = 0$
Since $x \neq y$,we have $x + y + xy = 0$
$y(1 + x) = -x$
$y = -\frac{x}{1 + x}$
Now,differentiate with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\frac{(1 + x)(1) - x(1)}{(1 + x)^2}$
$\frac{dy}{dx} = -\frac{1 + x - x}{(1 + x)^2}$
$\frac{dy}{dx} = -\frac{1}{(1 + x)^2} = -(1 + x)^{-2}$

(C) Given the equation: $\sin y = x \sin (a + y)$
We can express $x$ as: $x = \frac{\sin y}{\sin (a + y)}$
Differentiating both sides with respect to $x$ using the quotient rule:
$\frac{dx}{dx} = \frac{d}{dx} \left[ \frac{\sin y}{\sin (a + y)} \right]$
$1 = \frac{\cos y \cdot \frac{dy}{dx} \cdot \sin (a + y) - \sin y \cdot \cos (a + y) \cdot \frac{dy}{dx}}{\sin^2 (a + y)}$
Factor out $\frac{dy}{dx}$ in the numerator:
$1 = \frac{\frac{dy}{dx} [\sin (a + y) \cos y - \cos (a + y) \sin y]}{\sin^2 (a + y)}$
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$1 = \frac{\frac{dy}{dx} \cdot \sin (a + y - y)}{\sin^2 (a + y)}$
$1 = \frac{\frac{dy}{dx} \cdot \sin a}{\sin^2 (a + y)}$
Therefore,$\frac{dy}{dx} = \frac{\sin^2 (a + y)}{\sin a}$.

(B) Given the equation: $\tan (x + y) + \tan (x - y) = 1$
Differentiating both sides with respect to $x$ using the chain rule:
$\sec^2(x + y) \cdot \frac{d}{dx}(x + y) + \sec^2(x - y) \cdot \frac{d}{dx}(x - y) = 0$
$\sec^2(x + y) \left(1 + \frac{dy}{dx}\right) + \sec^2(x - y) \left(1 - \frac{dy}{dx}\right) = 0$
Expanding the terms:
$\sec^2(x + y) + \sec^2(x + y) \frac{dy}{dx} + \sec^2(x - y) - \sec^2(x - y) \frac{dy}{dx} = 0$
Grouping the $\frac{dy}{dx}$ terms:
$\frac{dy}{dx} \left( \sec^2(x + y) - \sec^2(x - y) \right) = - \left( \sec^2(x + y) + \sec^2(x - y) \right)$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = - \frac{\sec^2(x + y) + \sec^2(x - y)}{\sec^2(x + y) - \sec^2(x - y)}$
$\frac{dy}{dx} = \frac{\sec^2(x + y) + \sec^2(x - y)}{\sec^2(x - y) - \sec^2(x + y)}$
Thus,the correct option is $(b)$.

(C) Given the equation: $y \sec x + \tan x + x^2 y = 0$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(y \sec x) + \frac{d}{dx}(\tan x) + \frac{d}{dx}(x^2 y) = 0$
$\left( \frac{dy}{dx} \sec x + y \sec x \tan x \right) + \sec^2 x + \left( 2xy + x^2 \frac{dy}{dx} \right) = 0$
Grouping the terms containing $\frac{dy}{dx}$:
$\frac{dy}{dx} (\sec x + x^2) + y \sec x \tan x + \sec^2 x + 2xy = 0$
$\frac{dy}{dx} (\sec x + x^2) = -(2xy + \sec^2 x + y \sec x \tan x)$
$\frac{dy}{dx} = -\frac{2xy + \sec^2 x + y \sec x \tan x}{x^2 + \sec x}$.

(A) Given the equation: $\sin(xy) + \frac{x}{y} = x^2 - y$.
Differentiating both sides with respect to $x$:
$\cos(xy) \cdot (y + x \frac{dy}{dx}) + \frac{y(1) - x \frac{dy}{dx}}{y^2} = 2x - \frac{dy}{dx}$.
Multiply by $y^2$ to simplify:
$y^2 \cos(xy) (y + x \frac{dy}{dx}) + y - x \frac{dy}{dx} = 2xy^2 - y^2 \frac{dy}{dx}$.
Expand and group the $\frac{dy}{dx}$ terms:
$y^3 \cos(xy) + xy^2 \cos(xy) \frac{dy}{dx} + y - x \frac{dy}{dx} = 2xy^2 - y^2 \frac{dy}{dx}$.
$\frac{dy}{dx} (xy^2 \cos(xy) - x + y^2) = 2xy^2 - y - y^3 \cos(xy)$.
Therefore,$\frac{dy}{dx} = \frac{2xy^2 - y - y^3 \cos(xy)}{xy^2 \cos(xy) - x + y^2}$.
Factoring out $y$ from the numerator:
$\frac{dy}{dx} = \frac{y(2xy - 1 - y^2 \cos(xy))}{xy^2 \cos(xy) - x + y^2}$.

(B) Given equation: $\sin^2 x + 2\cos y + xy = 0$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(2\cos y) + \frac{d}{dx}(xy) = 0$
Using the chain rule and product rule:
$2\sin x \cos x - 2\sin y \frac{dy}{dx} + (y \cdot 1 + x \frac{dy}{dx}) = 0$
Using the identity $\sin 2x = 2\sin x \cos x$:
$\sin 2x - 2\sin y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx}(x - 2\sin y) = -(y + \sin 2x)$
$\frac{dy}{dx} = \frac{y + \sin 2x}{2\sin y - x}$

(A) Given the equation: ${x^3} + 8xy + {y^3} = 64$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^3) + \frac{d}{dx}(8xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(64)$
Applying the chain rule and product rule:
$3x^2 + 8(y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0$
Expanding the terms:
$3x^2 + 8y + 8x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$
Grouping the terms containing $\frac{dy}{dx}$:
$(8x + 3y^2)\frac{dy}{dx} = -(3x^2 + 8y)$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{3x^2 + 8y}{8x + 3y^2}$
Thus,the correct option is $A$.

(A) Given the equation: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(c) = 0$
Applying the chain rule and product rule:
$2ax + 2h(y + x\frac{dy}{dx}) + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$
Divide the entire equation by $2$:
$ax + hy + hx\frac{dy}{dx} + by\frac{dy}{dx} + g + f\frac{dy}{dx} = 0$
Grouping the terms containing $\frac{dy}{dx}$:
$\frac{dy}{dx}(hx + by + f) = -(ax + hy + g)$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f}$

(A) Given equations are:
$x^2 + y^2 = t - \frac{1}{t}$ --- $(1)$
$x^4 + y^4 = t^2 + \frac{1}{t^2}$ --- $(2)$
Squaring equation $(1)$,we get:
$(x^2 + y^2)^2 = (t - \frac{1}{t})^2$
$x^4 + y^4 + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2$
Substitute the value from equation $(2)$ into this expression:
$(t^2 + \frac{1}{t^2}) + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2$
$2x^2y^2 = -2$
$x^2y^2 = -1$
Since $x^2y^2 = -1$,we have $y^2 = -\frac{1}{x^2}$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = \frac{d}{dx}(-x^{-2})$
$2y \frac{dy}{dx} = -(-2)x^{-3}$
$2y \frac{dy}{dx} = \frac{2}{x^3}$
$y \frac{dy}{dx} = \frac{1}{x^3}$
Multiplying both sides by $x^3$,we get:
$x^3y \frac{dy}{dx} = 1$.

(A) Given the equation $\sin y = x \cos (a + y).$
We can express $x$ as $x = \frac{\sin y}{\cos (a + y)}.$
Differentiating both sides with respect to $y$ using the quotient rule $\frac{d}{dy} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dy} - u \frac{dv}{dy}}{v^2}$:
$\frac{dx}{dy} = \frac{\cos (a + y) \cdot \cos y - \sin y \cdot (-\sin (a + y))}{\cos^2 (a + y)}$
$\frac{dx}{dy} = \frac{\cos (a + y) \cos y + \sin y \sin (a + y)}{\cos^2 (a + y)}$
Using the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos (A - B)$:
$\frac{dx}{dy} = \frac{\cos (a + y - y)}{\cos^2 (a + y)} = \frac{\cos a}{\cos^2 (a + y)}$
Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{\cos^2 (a + y)}{\cos a}.$
Thus,the correct option is $A$.

(A) Given the equation $3\sin(xy) + 4\cos(xy) = 5$.
Let $u = xy$. Then $3\sin(u) + 4\cos(u) = 5$.
Differentiating both sides with respect to $x$:
$3\cos(u) \cdot \frac{du}{dx} - 4\sin(u) \cdot \frac{du}{dx} = 0$.
$(3\cos(u) - 4\sin(u)) \cdot \frac{du}{dx} = 0$.
Since $3\sin(u) + 4\cos(u) = 5$,the expression $3\cos(u) - 4\sin(u)$ cannot be zero for all $x$.
Thus,$\frac{du}{dx} = 0$.
Since $u = xy$,we have $\frac{d}{dx}(xy) = 0$.
Using the product rule: $y + x \cdot \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(C) Given the implicit function $f(x, y) = x^2 e^y + 2xy e^x + 13 = 0$.
To find $\frac{dy}{dx}$,we use the formula $\frac{dy}{dx} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$.
First,calculate the partial derivative with respect to $x$:
$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 e^y + 2xy e^x + 13) = 2x e^y + 2y e^x + 2xy e^x = 2x e^y + 2y e^x(1 + x)$.
Next,calculate the partial derivative with respect to $y$:
$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 e^y + 2xy e^x + 13) = x^2 e^y + 2x e^x$.
Now,substitute these into the formula:
$\frac{dy}{dx} = - \frac{2x e^y + 2y e^x(1 + x)}{x^2 e^y + 2x e^x}$.
Divide the numerator and the denominator by $x e^x$:
$\frac{dy}{dx} = - \frac{\frac{2x e^y}{e^x} + 2y(1 + x)}{\frac{x^2 e^y}{e^x} + 2x} = - \frac{2x e^{y-x} + 2y(x + 1)}{x(x e^{y-x} + 2)}$.
Thus,the correct option is $C$.

(D) Given the implicit function $\sin(x+y) = \log(x+y)$.
Let $u = x+y$. Then the equation becomes $\sin(u) = \log(u)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin(u)) = \frac{d}{dx}(\log(u))$
$\cos(u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$
Since $\frac{du}{dx} = \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}$,we have:
$\cos(u) \cdot (1 + \frac{dy}{dx}) = \frac{1}{u} \cdot (1 + \frac{dy}{dx})$
$(\cos(u) - \frac{1}{u}) \cdot (1 + \frac{dy}{dx}) = 0$
Since $\sin(u) = \log(u)$,$\cos(u) \neq \frac{1}{u}$ in general,therefore:
$1 + \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -1$.

(A) Given the equation $\ln (x + y) = 2xy$.
First,find the value of $y$ when $x = 0$:
$\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx} [\ln (x + y)] = \frac{d}{dx} [2xy]$
$\frac{1}{x + y} \cdot (1 + y') = 2(y + xy')$
Substitute $x = 0$ and $y = 1$ into the differentiated equation:
$\frac{1}{0 + 1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$
$1 \cdot (1 + y'(0)) = 2(1)$
$1 + y'(0) = 2$
$y'(0) = 2 - 1 = 1$.

(A) Given the equation ${x^y} = {e^{x - y}}$.
Taking the natural logarithm on both sides,we get:
$y \log x = x - y$
Rearranging the terms to solve for $y$:
$y(1 + \log x) = x$
$y = \frac{x}{1 + \log x}$
Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(1 + \log x)(1) - x(\frac{1}{x})}{(1 + \log x)^2}$
$\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2}$
$\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$
Since $1 + \log x = \log e + \log x = \log (ex)$:
$\frac{dy}{dx} = \log x \cdot [\log (ex)]^{-2}$.

(B) Given the equation: $(x - y)e^{x/(x - y)} = k$.
Taking the natural logarithm on both sides:
$\ln(x - y) + \frac{x}{x - y} = \ln k$.
Differentiating both sides with respect to $x$:
$\frac{1}{x - y}(1 - \frac{dy}{dx}) + \frac{(x - y)(1) - x(1 - \frac{dy}{dx})}{(x - y)^2} = 0$.
Multiply the entire equation by $(x - y)^2$:
$(x - y)(1 - \frac{dy}{dx}) + (x - y) - x(1 - \frac{dy}{dx}) = 0$.
$(x - y) - (x - y)\frac{dy}{dx} + x - y - x + x\frac{dy}{dx} = 0$.
Combine the terms involving $\frac{dy}{dx}$:
$(-x + y + x)\frac{dy}{dx} + (x - y + x - y - x) = 0$.
$y\frac{dy}{dx} + x - 2y = 0$.

(B) Given the equation: ${2^x} + {2^y} = {2^{x + y}}$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}({2^x}) + \frac{d}{dx}({2^y}) = \frac{d}{dx}({2^{x + y}})$
Using the chain rule,we get:
${2^x} \ln 2 + {2^y} \ln 2 \cdot \frac{dy}{dx} = {2^{x + y}} \ln 2 \cdot (1 + \frac{dy}{dx})$
Dividing by $\ln 2$:
${2^x} + {2^y} \frac{dy}{dx} = {2^{x + y}} + {2^{x + y}} \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
${2^y} \frac{dy}{dx} - {2^{x + y}} \frac{dy}{dx} = {2^{x + y}} - {2^x}$
$\frac{dy}{dx} ({2^y} - {2^{x + y}}) = {2^{x + y}} - {2^x}$
$\frac{dy}{dx} = \frac{{2^{x + y}} - {2^x}}{{2^y} - {2^{x + y}}}$
Since ${2^{x + y}} = {2^x} + {2^y}$,substitute this into the expression:
$\frac{dy}{dx} = \frac{({2^x} + {2^y}) - {2^x}}{{2^y} - ({2^x} + {2^y})} = \frac{{2^y}}{{2^y} - {2^x} - {2^y}} = \frac{{2^y}}{-{2^x}} = -{2^{y - x}}$
Wait,let's re-evaluate the simplification:
$\frac{dy}{dx} = \frac{{2^x}({2^y} - 1)}{{2^y}(1 - {2^x})} = {2^{x - y}} \frac{{2^y} - 1}{1 - {2^x}}$
Thus,the correct option is $B$.

(A) Given the equation: ${y^x} + {x^y} = {a^b}$.
Let $u = {x^y}$ and $v = {y^x}$.
Then $u + v = {a^b}$.
Differentiating with respect to $x$,we get $\frac{du}{dx} + \frac{dv}{dx} = 0$.
For $u = {x^y}$,taking log on both sides: $\log u = y \log x$.
Differentiating: $\frac{1}{u} \frac{du}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \implies \frac{du}{dx} = {x^y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) = y{x^{y - 1}} + {x^y} \log x \frac{dy}{dx}$.
For $v = {y^x}$,taking log on both sides: $\log v = x \log y$.
Differentiating: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1 \implies \frac{dv}{dx} = {y^x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = x{y^{x - 1}} \frac{dy}{dx} + {y^x} \log y$.
Substituting these into $\frac{du}{dx} + \frac{dv}{dx} = 0$:
$y{x^{y - 1}} + {x^y} \log x \frac{dy}{dx} + x{y^{x - 1}} \frac{dy}{dx} + {y^x} \log y = 0$.
$\frac{dy}{dx} ({x^y} \log x + x{y^{x - 1}}) = -(y{x^{y - 1}} + {y^x} \log y)$.
$\frac{dy}{dx} = - \frac{y{x^{y - 1}} + {y^x} \log y}{x{y^{x - 1}} + {x^y} \log x}$.

(C) Given the equation $y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \dots \infty}}}$.
Since the series is infinite,we can write $y = \sqrt{\log x + y}$.
Squaring both sides,we get $y^2 = \log x + y$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(\log x + y)$
$2y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x}$
$\frac{dy}{dx}(2y - 1) = \frac{1}{x}$
$\frac{dy}{dx} = \frac{1}{x(2y - 1)}$.
Thus,the correct option is $C$.

(B) Given the equation ${x^y} = {y^x}$.
Taking the natural logarithm on both sides,we get $y \log_e x = x \log_e y$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(y \log_e x) = \frac{d}{dx}(x \log_e y)$
$\frac{dy}{dx} \log_e x + y \cdot \frac{1}{x} = 1 \cdot \log_e y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \log_e x - \frac{x}{y} \frac{dy}{dx} = \log_e y - \frac{y}{x}$
$\frac{dy}{dx} \left( \log_e x - \frac{x}{y} \right) = \log_e y - \frac{y}{x}$
$\frac{dy}{dx} \left( \frac{y \log_e x - x}{y} \right) = \frac{x \log_e y - y}{x}$
$\frac{dy}{dx} = \frac{y(x \log_e y - y)}{x(y \log_e x - x)}$
Multiplying numerator and denominator by $-1$,we get $\frac{dy}{dx} = \frac{y(y - x \log_e y)}{x(x - y \log_e x)}$.

(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$,we can observe that the expression repeats itself after the first $x^2$.
Thus,we can write the equation as $y = x^2 + \frac{1}{y}$.
Multiplying both sides by $y$,we get $y^2 = x^2y + 1$.
Differentiating both sides with respect to $x$ using the product rule:
$2y \frac{dy}{dx} = x^2 \frac{dy}{dx} + y(2x)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2y \frac{dy}{dx} - x^2 \frac{dy}{dx} = 2xy$.
$\frac{dy}{dx}(2y - x^2) = 2xy$.
Therefore,$\frac{dy}{dx} = \frac{2xy}{2y - x^2}$.

(B) Given the equation: ${x^y} \cdot {y^x} = 1$.
Taking the natural logarithm on both sides: $\log({x^y} \cdot {y^x}) = \log(1)$.
Using logarithmic properties: $y \log x + x \log y = 0$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(y \log x) + \frac{d}{dx}(x \log y) = 0$.
Applying the product rule: $(y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}) + (x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y \cdot 1) = 0$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} (\log x + \frac{x}{y}) = - (\frac{y}{x} + \log y)$.
$\frac{dy}{dx} (\frac{y \log x + x}{y}) = - (\frac{y + x \log y}{x})$.
Therefore,$\frac{dy}{dx} = - \frac{y(y + x \log y)}{x(x + y \log x)}$.

(B) Given equation: ${2^x} + {2^y} = {2^{x + y}}$.
Differentiating both sides with respect to $x$:
${2^x} \ln(2) + {2^y} \ln(2) \frac{dy}{dx} = {2^{x + y}} \ln(2) \left( 1 + \frac{dy}{dx} \right)$.
Dividing by $\ln(2)$:
${2^x} + {2^y} \frac{dy}{dx} = {2^{x + y}} + {2^{x + y}} \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} ({2^y} - {2^{x + y}}) = {2^{x + y}} - {2^x}$.
Thus,$\frac{dy}{dx} = \frac{{2^{x + y}} - {2^x}}{{2^y} - {2^{x + y}}}$.
At $x = 1$ and $y = 1$:
$\frac{dy}{dx} = \frac{{2^{1 + 1}} - {2^1}}{{2^1} - {2^{1 + 1}}} = \frac{{2^2} - 2}{2 - {2^2}} = \frac{4 - 2}{2 - 4} = \frac{2}{-2} = -1$.

(D) Given $f''(x) - g''(x) = 0$.
Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.
At $x = 1$,$f'(1) - g'(1) = c \implies 2 - 4 = c \implies c = -2$.
Thus,$f'(x) - g'(x) = -2$.
Integrating again with respect to $x$,we get $f(x) - g(x) = -2x + c_1$,where $c_1$ is a constant.
At $x = 2$,$f(2) - g(2) = -2(2) + c_1 \implies 3 - 9 = -4 + c_1 \implies -6 = -4 + c_1 \implies c_1 = -2$.
Therefore,$f(x) - g(x) = -2x - 2$.
At $x = 3/2$,$f(3/2) - g(3/2) = -2(3/2) - 2 = -3 - 2 = -5$.

(D) Given the equation $2t = v^2$.
To find $\frac{dv}{dt}$,we differentiate both sides with respect to $t$.
$\frac{d}{dt}(2t) = \frac{d}{dt}(v^2)$
$2 = 2v \cdot \frac{dv}{dt}$
Dividing both sides by $2v$,we get:
$\frac{dv}{dt} = \frac{2}{2v} = \frac{1}{v}$.

(C) Given the equation $x{e^{xy}} = y + {\sin ^2}x$.
When $x = 0$,substituting into the equation gives $0 \cdot {e^0} = y + {\sin ^2}(0)$,which implies $y = 0$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(x{e^{xy}}) = \frac{d}{dx}(y + {\sin ^2}x)$
${e^{xy}} + x{e^{xy}} \cdot \frac{d}{dx}(xy) = \frac{dy}{dx} + 2\sin x \cos x$
${e^{xy}} + x{e^{xy}} \left( y + x \frac{dy}{dx} \right) = \frac{dy}{dx} + \sin(2x)$
Substituting $x = 0$ and $y = 0$ into the differentiated equation:
${e^0} + 0 \cdot {e^0} (0 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + \sin(0)$
$1 + 0 = \frac{dy}{dx} + 0$
Therefore,$\frac{dy}{dx} = 1$.

(B) Given the infinite power tower $y = x^{x^{x^{\dots\infty}}}$.
Since the exponent repeats infinitely,we can write $y = x^y$.
Taking the natural logarithm on both sides,we get $\log y = y \log x$.
Differentiating both sides with respect to $x$ using the product rule on the right side:
$\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{dy}{dx}$.
$\frac{1}{y} \frac{dy}{dx} = \frac{y}{x} + \log x \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} (\frac{1}{y} - \log x) = \frac{y}{x}$.
$\frac{dy}{dx} (\frac{1 - y \log x}{y}) = \frac{y}{x}$.
Therefore,$\frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)}$.

(B) Given the equation: ${x^2} + {y^2} = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}({x^2} + {y^2}) = \frac{d}{dx}(1)$
$2x + 2y \frac{dy}{dx} = 0$
$x + y y' = 0$
Differentiating again with respect to $x$ using the product rule on $y y'$:
$\frac{d}{dx}(x) + \frac{d}{dx}(y y') = 0$
$1 + (y \cdot y'' + y' \cdot y') = 0$
$1 + y y'' + (y')^2 = 0$
Thus,the relation is $yy'' + (y')^2 + 1 = 0$.

(B) Given the equation: $y \cos x + x \cos y = \pi$.
At $x = 0$,we have $y \cos(0) + 0 \cos y = \pi$,which implies $y = \pi$.
Differentiating both sides with respect to $x$:
$-y \sin x + y' \cos x + \cos y - x \sin y \cdot y' = 0$.
At $x = 0$ and $y = \pi$:
$-(\pi) \sin(0) + y'(0) \cos(0) + \cos(\pi) - 0 \sin(\pi) \cdot y'(0) = 0$.
$0 + y'(0) \cdot 1 - 1 - 0 = 0 \implies y'(0) = 1$.
Now,differentiate the first derivative expression again:
$-y' \sin x - y \cos x + y'' \cos x - y' \sin x - \sin y \cdot y' - x(\cos y \cdot (y')^2 + \sin y \cdot y'') - \sin y \cdot y' = 0$.
Substitute $x = 0, y = \pi, y' = 1$:
$-1 \cdot 0 - \pi \cdot 1 + y''(0) \cdot 1 - 1 \cdot 0 - 0 - 0 - 0 = 0$.
$-\pi + y''(0) = 0 \implies y''(0) = \pi$.

(A) Given the curve equation: $xy + ax - by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$ into the equation:
$(1)(1) + a(1) - b(1) = 0
\Rightarrow 1 + a - b = 0
\Rightarrow a - b = -1$ (Equation $1$).
Now,differentiate the equation $xy + ax - by = 0$ with respect to $x$:
$x \frac{dy}{dx} + y + a - b \frac{dy}{dx} = 0$.
Rearranging for $\frac{dy}{dx}$:
$\frac{dy}{dx} (x - b) = -(y + a)
\Rightarrow \frac{dy}{dx} = -\frac{y + a}{x - b}$.
Given that the slope at $(1, 1)$ is $2$:
$2 = -\frac{1 + a}{1 - b}
\Rightarrow 2(1 - b) = -(1 + a)
\Rightarrow 2 - 2b = -1 - a
\Rightarrow a - 2b = -3$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$:
$(a - b) - (a - 2b) = -1 - (-3)
\Rightarrow a - b - a + 2b = -1 + 3
\Rightarrow b = 2$.
Substituting $b = 2$ into Equation $1$:
$a - 2 = -1
\Rightarrow a = 1$.
Thus,$a = 1$ and $b = 2$.

(A) Given the curve $y = \cos(x + y)$.
Taking the derivative with respect to $x$:
$\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$
$\frac{dy}{dx} (1 + \sin(x + y)) = -\sin(x + y)$
$\frac{dy}{dx} = -\frac{\sin(x + y)}{1 + \sin(x + y)}$
Since the tangent is parallel to $x + 2y = 0$,its slope is $-\frac{1}{2}$.
Setting the derivative equal to $-\frac{1}{2}$:
$-\frac{\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2}$
$2\sin(x + y) = 1 + \sin(x + y)$
$\sin(x + y) = 1$
This implies $x + y = \frac{\pi}{2} + 2n\pi$.
Substituting this into the original equation $y = \cos(x + y)$:
$y = \cos(\frac{\pi}{2} + 2n\pi) = 0$.
Since $y = 0$,then $x + 0 = \frac{\pi}{2} \implies x = \frac{\pi}{2}$.
Thus,the point is $(\frac{\pi}{2}, 0)$.

(D) Given the equation: ${x^{2x}} - 2{x^x}\cot y - 1 = 0$ ........$(i)$
At $x = 1$,we have:
$1^{2(1)} - 2(1^1)\cot y - 1 = 0$
$1 - 2\cot y - 1 = 0$
$\Rightarrow \cot y = 0 \quad \therefore \quad y = \frac{\pi}{2}$
Differentiating $(i)$ with respect to $x$:
$\frac{d}{dx}(x^{2x}) - 2 \frac{d}{dx}(x^x \cot y) = 0$
$x^{2x} \cdot \frac{d}{dx}(2x \ln x) - 2 \left[ x^x \ln x \cdot \cot y + x^x \cdot (-\csc^2 y) \frac{dy}{dx} + \cot y \cdot x^x(1 + \ln x) \right] = 0$
Using $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$:
$2x^{2x}(1 + \ln x) - 2 \left[ x^x(1 + \ln x) \cot y - x^x \csc^2 y \frac{dy}{dx} \right] = 0$
At $P(1, \frac{\pi}{2})$,we have $x = 1, y = \frac{\pi}{2}, \cot y = 0, \csc^2 y = 1, \ln 1 = 0$:
$2(1)^{2}(1 + 0) - 2 \left[ 1(1 + 0)(0) - 1(1) \left(\frac{dy}{dx}\right)_{P} \right] = 0$
$2 - 2 \left[ -\left(\frac{dy}{dx}\right)_{P} \right] = 0$
$2 + 2 \left(\frac{dy}{dx}\right)_{P} = 0$
$\left(\frac{dy}{dx}\right)_{P} = -1$

(D) Given $f(x) = |\log 2 - \sin x|$. Since $\log 2 \approx 0.693 < 1$,for $x$ near $0$,$\sin x$ is small,so $\log 2 - \sin x > 0$. Thus,$f(x) = \log 2 - \sin x$ in the neighborhood of $x=0$.
Then $g(x) = f(f(x)) = \log 2 - \sin(f(x)) = \log 2 - \sin(\log 2 - \sin x)$.
Since $f(x)$ is differentiable at $x=0$ and the composition of differentiable functions is differentiable,$g(x)$ is differentiable at $x=0$.
Now,$g'(x) = -\cos(\log 2 - \sin x) \cdot (-\cos x) = \cos(\log 2 - \sin x) \cdot \cos x$.
Evaluating at $x=0$,$g'(0) = \cos(\log 2 - \sin 0) \cdot \cos 0 = \cos(\log 2) \cdot 1 = \cos(\log 2)$.

(B) Given: ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$.
We know that ${({x^2} + {y^2})^2} = {x^4} + {y^4} + 2{x^2}{y^2}$.
Substituting the given values: ${\left( {t - \frac{1}{t}} \right)^2} = {t^2} + \frac{1}{{{t^2}}} + 2{x^2}{y^2}$.
Expanding the left side: ${t^2} - 2 + \frac{1}{{{t^2}}} = {t^2} + \frac{1}{{{t^2}}} + 2{x^2}{y^2}$.
This simplifies to $-2 = 2{x^2}{y^2}$,which means ${x^2}{y^2} = -1$,or ${y^2} = -\frac{1}{{{x^2}}}$.
Differentiating both sides with respect to $x$: $2y \frac{{dy}}{{dx}} = -\frac{d}{{dx}}({x^{-2}}) = -(-2){x^{-3}} = \frac{2}{{{x^3}}}$.
Therefore,$\frac{{dy}}{{dx}} = \frac{2}{{{x^3} \cdot 2y}} = \frac{1}{{{x^3}y}}$.

(B) Given equation: $\sin(xy) + \cos(xy) = 0$
This implies $\sin(xy) = -\cos(xy)$,which means $\tan(xy) = -1$.
Since $\tan(xy) = -1$,we have $xy = n\pi - \frac{\pi}{4}$ for some integer $n$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(xy) = \frac{d}{dx}(n\pi - \frac{\pi}{4})$
$y + x \frac{dy}{dx} = 0$
$x \frac{dy}{dx} = -y$
$\frac{dy}{dx} = -\frac{y}{x}$

(D) Given the equation: $y^2e^{xy} = 9e^{-3}x^2$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(y^2)e^{xy} + y^2\frac{d}{dx}(e^{xy}) = 9e^{-3}\frac{d}{dx}(x^2)$
$2y\frac{dy}{dx}e^{xy} + y^2e^{xy}\left(y + x\frac{dy}{dx}\right) = 18e^{-3}x$
Substitute $x = -1$ and $y = 3$:
$2(3)\frac{dy}{dx}e^{-3} + (3)^2e^{-3}\left(3 + (-1)\frac{dy}{dx}\right) = 18e^{-3}(-1)$
Divide the entire equation by $e^{-3}$ (since $e^{-3} \neq 0$):
$6\frac{dy}{dx} + 9(3 - \frac{dy}{dx}) = -18$
$6\frac{dy}{dx} + 27 - 9\frac{dy}{dx} = -18$
$-3\frac{dy}{dx} = -18 - 27$
$-3\frac{dy}{dx} = -45$
$\frac{dy}{dx} = 15$.

(A) Given the equation: $2^x + 2^y = 2^{x + y}$.
Divide both sides by $2^{x+y}$:
$\frac{2^x}{2^{x+y}} + \frac{2^y}{2^{x+y}} = 1$
$2^{-y} + 2^{-x} = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2^{-y}) + \frac{d}{dx}(2^{-x}) = \frac{d}{dx}(1)$
Using the chain rule,$\frac{d}{dx}(a^u) = a^u \ln(a) \frac{du}{dx}$:
$2^{-y} \ln(2) \cdot (-\frac{dy}{dx}) + 2^{-x} \ln(2) \cdot (-1) = 0$.
Divide by $-\ln(2)$:
$2^{-y} \frac{dy}{dx} + 2^{-x} = 0$
$2^{-y} \frac{dy}{dx} = -2^{-x}$
$\frac{dy}{dx} = -\frac{2^{-x}}{2^{-y}} = -\frac{2^y}{2^x}$.
Thus,option $A$ is correct.

(C) Given the equation of the curve: $y - e^{xy} + x = 0$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} - e^{xy} \cdot (y + x \frac{dy}{dx}) + 1 = 0$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} (1 - x e^{xy}) = y e^{xy} - 1$.
$\frac{dy}{dx} = \frac{y e^{xy} - 1}{1 - x e^{xy}}$.
$A$ vertical tangent occurs where the denominator is zero and the numerator is non-zero,i.e.,$\frac{dy}{dx} \to \infty$.
Setting the denominator to zero: $1 - x e^{xy} = 0$,which implies $x e^{xy} = 1$.
From the original equation $y - e^{xy} + x = 0$,we have $e^{xy} = x + y$.
Substituting $e^{xy} = x + y$ into $x e^{xy} = 1$,we get $x(x + y) = 1$,or $x^2 + xy = 1$.
However,we must check if any point $(x, y)$ satisfies both $e^{xy} = x + y$ and $x e^{xy} = 1$.
If $x = 0$,then $1 - 0 = 0$,which is $1 = 0$,a contradiction.
If $x \neq 0$,then $e^{xy} = \frac{1}{x}$.
Substituting this into the original equation: $y - \frac{1}{x} + x = 0 \implies y = \frac{1}{x} - x = \frac{1 - x^2}{x}$.
Now,$e^{xy} = e^{x(\frac{1-x^2}{x})} = e^{1-x^2}$.
We need $e^{1-x^2} = \frac{1}{x}$.
Testing the given options:
For $(1, 0)$: $0 - e^0 + 1 = 0 \implies 0 - 1 + 1 = 0$ (Satisfied).
At $(1, 0)$,the denominator $1 - x e^{xy} = 1 - 1 \cdot e^0 = 1 - 1 = 0$.
The numerator $y e^{xy} - 1 = 0 \cdot e^0 - 1 = -1 \neq 0$.
Since the denominator is $0$ and the numerator is non-zero,the slope is undefined (vertical tangent).
Thus,the curve has a vertical tangent at $(1, 0)$.

(D) Given $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ and $f'(0) = 1$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$f''(x) = f^{-1}(f(x)) \cdot f'(x)$.
Since $f(x)$ is invertible,$f^{-1}(f(x)) = x$.
Thus,$f''(x) = x f'(x)$.
Rearranging the terms,we get $\frac{f''(x)}{f'(x)} = x$.
Integrating both sides with respect to $x$:
$\ln|f'(x)| = \frac{x^2}{2} + C$.
At $x = 0$,$f'(0) = 1$,so $\ln|1| = 0 + C$,which implies $C = 0$.
Therefore,$\ln|f'(x)| = \frac{x^2}{2}$,which gives $f'(x) = e^{x^2/2}$.
For $x = 1$,$f'(1) = e^{1/2} = \sqrt{e}$.