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.