Differentiation of implicit function Questions in English

(A) Let $L = \lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$.
Evaluating the integral, we get:
$L = \lim _{x \rightarrow 1} \frac{1}{x-1} [t^2]_4^{f(x)} = \lim _{x \rightarrow 1} \frac{[f(x)]^2 - 16}{x-1}$.
Since $f(1) = 4$, the expression is in the $\frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx} ([f(x)]^2 - 16)}{\frac{d}{dx} (x-1)} = \lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}$.
Substituting the values $f(1) = 4$ and $f^{\prime}(1) = 2$:
$L = 2 \times f(1) \times f^{\prime}(1) = 2 \times 4 \times 2 = 16$.

(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.
Setting $y=0$, we get $\frac{f(x)}{f(0)}=f(x)$, which implies $f(0)=1$.
Differentiating both sides with respect to $x$, we get $\frac{f^{\prime}(x)}{f(y)}=f^{\prime}(x-y)$.
Setting $x=0$, we have $\frac{f^{\prime}(0)}{f(y)}=f^{\prime}(-y)$.
Since $f^{\prime}(0)=p$, we get $f^{\prime}(-y) = \frac{p}{f(y)}$.
Also, differentiating the original equation with respect to $y$, we get $f(x) \cdot (-\frac{f^{\prime}(y)}{(f(y))^2}) = f^{\prime}(x-y) \cdot (-1)$.
This simplifies to $\frac{f(x) f^{\prime}(y)}{(f(y))^2} = f^{\prime}(x-y)$.
At $y=0$, $\frac{f(x) f^{\prime}(0)}{(f(0))^2} = f^{\prime}(x)$, so $f^{\prime}(x) = p f(x)$.
This is a linear differential equation with solution $f(x) = e^{px}$.
Then $f^{\prime}(x) = p e^{px}$.
Given $f^{\prime}(5) = q$, we have $p e^{5p} = q$, so $e^{5p} = \frac{q}{p}$.
We need $f^{\prime}(-5) = p e^{-5p} = \frac{p}{e^{5p}} = \frac{p}{q/p} = \frac{p^2}{q}$.

(B) Given the equation $x^2+y^2=4$.
Differentiating both sides with respect to $x$, we get:
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)$
$2x + 2y \frac{dy}{dx} = 0$
Dividing the entire equation by $2$, we get:
$x + y \frac{dy}{dx} = 0$
Therefore, $y \frac{dy}{dx} + x = 0$.

(B) Given $y = \frac{1}{1 + x + \ln x}$.
Taking the reciprocal, we get $\frac{1}{y} = 1 + x + \ln x$.
Differentiating both sides with respect to $x$:
$-\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.
From the original equation, $1 + \ln x = \frac{1}{y} - x$.
Substitute this into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.
Alternatively, differentiating $y(1 + x + \ln x) = 1$ with respect to $x$:
$\frac{dy}{dx}(1 + x + \ln x) + y(1 + \frac{1}{x}) = 0$.
Since $1 + x + \ln x = \frac{1}{y}$, we have:
$\frac{dy}{dx} \cdot \frac{1}{y} + y \left(\frac{x + 1}{x}\right) = 0$.
$\frac{1}{y} \frac{dy}{dx} = -y \left(\frac{x + 1}{x}\right)$.
$x \frac{dy}{dx} = -y^{2}(x + 1)$.
Wait, let us re-evaluate: $\frac{1}{y} = 1 + x + \ln x \implies \frac{d}{dx}(\frac{1}{y}) = \frac{d}{dx}(1 + x + \ln x) \implies -\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.
$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.
Checking option $B$: $x \frac{dy}{dx} = y(y \ln x - 1)$.
If $y = \frac{1}{1 + x + \ln x}$, then $y \ln x - 1 = y \ln x - y(1 + x + \ln x) = y(\ln x - 1 - x - \ln x) = y(-1 - x) = -y(1 + x)$.
So $y(y \ln x - 1) = y(-y(1 + x)) = -y^{2}(1 + x)$.
Thus, $x \frac{dy}{dx} = -y^{2}(1 + x)$, which matches.

(C) Given $y = \sqrt{\cos x^2 + y}$.
Squaring both sides, we get $y^2 = \cos x^2 + y$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = -\sin x^2 \cdot (2x) + \frac{dy}{dx}$.
Rearranging the terms:
$(2y - 1) \frac{dy}{dx} = -2x \sin x^2$.
Thus, $\frac{dy}{dx} = \frac{-2x \sin x^2}{2y - 1}$.
Comparing with $\frac{dy}{dx} = \frac{f(x)}{2y - 1}$, we get $f(x) = -2x \sin x^2$.
Now, $\int f(x) dx = \int -2x \sin x^2 dx$.
Let $u = x^2$, then $du = 2x dx$.
Substituting these, we get $\int -\sin u du = \cos u + c$.
Substituting back $u = x^2$, we get $\cos x^2 + c$.

(C) Step $1$: Since $(1, 1)$ lies on the curve $xy + ax + by = 0$, we have $(1)(1) + a(1) + b(1) = 0$, which implies $1 + a + b = 0$, or $a + b = -1$.
Step $2$: Differentiate the equation $xy + ax + by = 0$ with respect to $x$: $y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$.
Step $3$: At $(1, 1)$, the slope $\frac{dy}{dx} = \tan(\tan^{-1} 2) = 2$. Substituting these values: $1 + (1)(2) + a + b(2) = 0$.
Step $4$: This simplifies to $1 + 2 + a + 2b = 0$, so $a + 2b = -3$.
Step $5$: Solving the system $a + b = -1$ and $a + 2b = -3$, subtract the first from the second: $(a + 2b) - (a + b) = -3 - (-1) \implies b = -2$. Then $a = -1 - (-2) = 1$.
Step $6$: Calculate $\frac{ab}{a + b} = \frac{(1)(-2)}{-1} = \frac{-2}{-1} = 2$.

(A) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides: $(y + x) + (y - x) + 2\sqrt{y^2 - x^2} = c^2 \implies 2y + 2\sqrt{y^2 - x^2} = c^2 \implies \sqrt{y^2 - x^2} = \frac{c^2}{2} - y$.
Squaring again: $y^2 - x^2 = \frac{c^4}{4} - c^2y + y^2 \implies -x^2 = \frac{c^4}{4} - c^2y \implies c^2y = x^2 + \frac{c^4}{4}$.
Differentiating with respect to $x$: $c^2 \frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{2x}{c^2}$.
From $\sqrt{y^2 - x^2} = \frac{c^2}{2} - y$, we have $\frac{c^2}{2} = y + \sqrt{y^2 - x^2}$.
Thus, $\frac{dy}{dx} = \frac{2x}{2(y + \sqrt{y^2 - x^2})} = \frac{x}{y + \sqrt{y^2 - x^2}} = \frac{x}{y + y\sqrt{1 - (x/y)^2}} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$.
Rationalizing the denominator: $\frac{dy}{dx} = \frac{x/y (1 - \sqrt{1 - (x/y)^2})}{1 - (1 - (x/y)^2)} = \frac{x/y (1 - \sqrt{1 - (x/y)^2})}{(x/y)^2} = \frac{1 - \sqrt{1 - (x/y)^2}}{x/y} = \frac{y}{x} - \sqrt{(y/x)^2 - 1}$.
Comparing with $\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$, we get $f(x) = \frac{y}{x}$.

(B) Step $1$: Find $y$ at $x = 0$. Substituting $x = 0$ into $e^y + xy = e$, we get $e^y + 0 = e$, so $y = 1$.
Step $2$: Differentiate $e^y + xy = e$ with respect to $x$: $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$.
Step $3$: At $x = 0, y = 1$, substitute these into the derivative: $e^1 \frac{dy}{dx} + 1 + 0 = 0 \implies \frac{dy}{dx} = -\frac{1}{e}$.
Step $4$: Differentiate $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ again: $e^y (\frac{dy}{dx})^2 + e^y \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$.
Step $5$: Substitute $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$ into the second derivative equation: $e^1(-\frac{1}{e})^2 + e^1 \frac{d^2y}{dx^2} + 2(-\frac{1}{e}) + 0 = 0$.
Step $6$: Simplify: $\frac{1}{e} + e \frac{d^2y}{dx^2} - \frac{2}{e} = 0 \implies e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$.
Step $7$: The ordered pair is $(-\frac{1}{e}, \frac{1}{e^2})$.

(A) Given $y^m + y^{-m} = 2x$.
Differentiating both sides with respect to $x$:
$m y^{m-1} \frac{dy}{dx} - m y^{-m-1} \frac{dy}{dx} = 2$.
$\frac{dy}{dx} (m y^{m-1} - m y^{-m-1}) = 2$.
$\frac{dy}{dx} = \frac{2}{m(y^{m-1} - y^{-m-1})} = \frac{2y}{m(y^m - y^{-m})}$.
From the given equation, $(y^m - y^{-m})^2 = (y^m + y^{-m})^2 - 4 = (2x)^2 - 4 = 4(x^2 - 1)$.
So, $y^m - y^{-m} = 2\sqrt{x^2 - 1}$.
Substituting this into the derivative: $\frac{dy}{dx} = \frac{2y}{m(2\sqrt{x^2 - 1})} = \frac{y}{m\sqrt{x^2 - 1}}$.
Squaring both sides: $\left( \frac{dy}{dx} \right)^2 = \frac{y^2}{m^2(x^2 - 1)}$.
Therefore, $(x^2 - 1) \left( \frac{dy}{dx} \right)^2 = \frac{y^2}{m^2}$.

(B) Given equation: $x e^{xy} = y + \sin^2 x$.
Step $1$: Find the value of $y$ at $x = 0$.
Substitute $x = 0$ into the equation: $0 \cdot e^{0} = y + \sin^2(0) \implies 0 = y + 0 \implies y = 0$.
Step $2$: Differentiate both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(x e^{xy}) = \frac{d}{dx}(y + \sin^2 x)$
$e^{xy} + x \cdot e^{xy} \cdot (y + x \frac{dy}{dx}) = \frac{dy}{dx} + 2 \sin x \cos x$.
Step $3$: Substitute $x = 0$ and $y = 0$ into the differentiated equation:
$e^{0} + 0 \cdot e^{0} \cdot (0 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + 2 \sin(0) \cos(0)$
$1 + 0 = \frac{dy}{dx} + 0$
$\frac{dy}{dx} = 1$.

(C) Given equation: $3y^2 - 2xy - x = 0$.
First, find the value of $x$ when $y = 2$: $3(2)^2 - 2x(2) - x = 0 \implies 12 - 4x - x = 0 \implies 12 = 5x \implies x = \frac{12}{5}$.
Differentiate the equation with respect to $x$: $\frac{d}{dx}(3y^2) - \frac{d}{dx}(2xy) - \frac{d}{dx}(x) = 0$.
$6y \frac{dy}{dx} - (2y + 2x \frac{dy}{dx}) - 1 = 0$.
Substitute $y = 2$ and $x = \frac{12}{5}$: $6(2) \frac{dy}{dx} - 2(2) - 2(\frac{12}{5}) \frac{dy}{dx} - 1 = 0$.
$12 \frac{dy}{dx} - 4 - \frac{24}{5} \frac{dy}{dx} - 1 = 0$.
$(12 - \frac{24}{5}) \frac{dy}{dx} = 5$.
$(\frac{60 - 24}{5}) \frac{dy}{dx} = 5$.
$\frac{36}{5} \frac{dy}{dx} = 5$.
$\frac{dy}{dx} = \frac{25}{36}$.

(D) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides:
$(y + x) + (y - x) + 2\sqrt{(y + x)(y - x)} = c^2$
$2y + 2\sqrt{y^2 - x^2} = c^2$
$\sqrt{y^2 - x^2} = \frac{c^2 - 2y}{2} = \frac{c^2}{2} - y$
Squaring again:
$y^2 - x^2 = (\frac{c^2}{2})^2 - c^2y + y^2$
$-x^2 = \frac{c^4}{4} - c^2y$
$c^2y = x^2 + \frac{c^4}{4}$
Differentiating with respect to $x$:
$c^2 \frac{dy}{dx} = 2x$
$\frac{dy}{dx} = \frac{2x}{c^2}$
From $\sqrt{y^2 - x^2} = \frac{c^2}{2} - y$, we have $c^2 = 2(\sqrt{y^2 - x^2} + y)$.
Substitute $c^2$ into $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{2x}{2(\sqrt{y^2 - x^2} + y)} = \frac{x}{\sqrt{y^2 - x^2} + y} = \frac{x/y}{\sqrt{(y^2 - x^2)/y^2} + 1} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$
This does not match the form directly. Let's re-evaluate $\frac{dy}{dx}$ from $y^2 - x^2 = (\frac{c^2}{2} - y)^2$:
$2y \frac{dy}{dx} - 2x = 2(\frac{c^2}{2} - y)(-\frac{dy}{dx})$
$y \frac{dy}{dx} - x = -(\frac{c^2}{2} - y) \frac{dy}{dx} = -\sqrt{y^2 - x^2} \frac{dy}{dx}$
$\frac{dy}{dx} (y + \sqrt{y^2 - x^2}) = x$
$\frac{dy}{dx} = \frac{x}{y + \sqrt{y^2 - x^2}} = \frac{x}{y + y\sqrt{1 - (x/y)^2}} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$.
Actually, the expression $\frac{dy}{dx} = \frac{x}{y + \sqrt{y^2 - x^2}}$ can be rationalized as $\frac{x(y - \sqrt{y^2 - x^2})}{y^2 - (y^2 - x^2)} = \frac{x(y - \sqrt{y^2 - x^2})}{x^2} = \frac{y - \sqrt{y^2 - x^2}}{x} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}$.
Comparing with $\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2} - 1}$, we get $K = \frac{y}{x}$.

(B) Given: $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$.
Multiply by $\sqrt{xy}$: $x + y = 6\sqrt{xy}$.
Square both sides: $(x + y)^2 = 36xy$.
$x^2 + 2xy + y^2 = 36xy \implies x^2 - 34xy + y^2 = 0$.
Differentiate with respect to $x$: $\frac{d}{dx}(x^2) - 34\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$.
$2x - 34(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$.
Divide by $2$: $x - 17y - 17x\frac{dy}{dx} + y\frac{dy}{dx} = 0$.
$x - 17y = \frac{dy}{dx}(17x - y)$.
$\frac{dy}{dx} = \frac{x - 17y}{17x - y}$.

(B) Given $y = (\tan x + y)^{1/3}$.
Cube both sides: $y^3 = \tan x + y$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^3) = \frac{d}{dx}(\tan x + y)$.
$3y^2 \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$.
Rearranging the terms: $3y^2 \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x$.
$\frac{dy}{dx}(3y^2 - 1) = \sec^2 x$.
Therefore, $\frac{dy}{dx} = \frac{\sec^2 x}{3y^2 - 1}$.

(C) Given $x^{1/2} y^{1/3} = (x + y)^n$.
Taking natural logarithm on both sides: $\frac{1}{2} \ln x + \frac{1}{3} \ln y = n \ln(x + y)$.
Differentiating with respect to $x$: $\frac{1}{2x} + \frac{1}{3y} \frac{dy}{dx} = \frac{n}{x + y} (1 + \frac{dy}{dx})$.
Given $x \frac{dy}{dx} - y = 0 \implies \frac{dy}{dx} = \frac{y}{x}$.
Substituting $\frac{dy}{dx} = \frac{y}{x}$ into the differentiated equation: $\frac{1}{2x} + \frac{1}{3y} (\frac{y}{x}) = \frac{n}{x + y} (1 + \frac{y}{x})$.
$\frac{1}{2x} + \frac{1}{3x} = \frac{n}{x + y} (\frac{x + y}{x})$.
$\frac{3 + 2}{6x} = \frac{n}{x}$.
$\frac{5}{6x} = \frac{n}{x} \implies n = \frac{5}{6}$.