Differentiation of infinite series Questions in English

(A) Given the infinite series $y = e^{x + e^{x + e^{x + \dots \infty}}}$.
Since the exponent repeats,we can write the equation as $y = e^{x + y}$.
Taking the natural logarithm on both sides,we get $\ln(y) = x + y$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = 1 + \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$,we have $\frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} = 1$.
$\frac{dy}{dx} (\frac{1}{y} - 1) = 1$.
$\frac{dy}{dx} (\frac{1 - y}{y}) = 1$.
Therefore,$\frac{dy}{dx} = \frac{y}{1 - y}$.

(A) Given $y = (\sin x)^{(\sin x)^{(\sin x)^{...\infty}}}$.
Since the exponent is an infinite series,we can write $y = (\sin x)^y$.
Taking the natural logarithm on both sides,we get $\ln y = y \ln(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x$.
$\frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \ln(\sin x) = y \cot x$.
$\frac{dy}{dx} \left( \frac{1}{y} - \ln(\sin x) \right) = y \cot x$.
$\frac{dy}{dx} \left( \frac{1 - y \ln(\sin x)}{y} \right) = y \cot x$.
Therefore,$\frac{dy}{dx} = \frac{y^2 \cot x}{1 - y \ln(\sin x)}$.

(D) Given $y = (\sqrt{x})^{(\sqrt{x})^{(\sqrt{x})^{\dots\infty}}}$.
Since the exponent is infinite,we can write $y = (\sqrt{x})^y$.
Taking the natural logarithm on both sides: $\log y = \log((\sqrt{x})^y) = y \log(\sqrt{x}) = y \log(x^{1/2}) = \frac{1}{2} y \log x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \log x \cdot \frac{dy}{dx} + y \cdot \frac{1}{x} \right)$.
Multiply by $2y$ to clear the fractions:
$2 \frac{dy}{dx} = y \log x \frac{dy}{dx} + \frac{y^2}{x}$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} (2 - y \log x) = \frac{y^2}{x}$.
Therefore,$\frac{dy}{dx} = \frac{y^2}{x(2 - y \log x)}$.
Comparing this result with the given options,none of the options $A, B,$ or $C$ match the derived expression. Thus,the correct option is $D$.

(D) Given the infinite series $y = \sqrt {x + \sqrt {x + \sqrt {x + \dots \infty } } }$.
Since the series is infinite,we can write it as $y = \sqrt {x + y}$.
Squaring both sides,we get $y^2 = x + y$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(x + y)$.
Using the chain rule,$2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = 1$.
$\frac{dy}{dx}(2y - 1) = 1$.
Therefore,$\frac{dy}{dx} = \frac{1}{2y - 1}$.

(C) Given the equation $x = e^{y + e^{y + \dots \infty}}$.
Since the exponent is an infinite series,we can write it as $x = e^{y + x}$.
Taking the natural logarithm on both sides,we get $\ln(x) = y + x$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(\ln(x)) = \frac{d}{dx}(y + x)$
$\frac{1}{x} = \frac{dy}{dx} + 1$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{1}{x} - 1 = \frac{1 - x}{x}$.

(A) Given the infinite continued fraction: $y = \frac{x}{a + \frac{x}{b + y}}$.
Multiply both sides by $(a + \frac{x}{b + y})$: $y(a + \frac{x}{b + y}) = x$.
$ay + \frac{xy}{b + y} = x$.
Multiply by $(b + y)$: $ay(b + y) + xy = x(b + y)$.
$aby + ay^2 + xy = bx + xy$.
$aby + ay^2 = bx$.
Now,differentiate both sides with respect to $x$: $\frac{d}{dx}(aby + ay^2) = \frac{d}{dx}(bx)$.
$ab \frac{dy}{dx} + 2ay \frac{dy}{dx} = b$.
$\frac{dy}{dx}(ab + 2ay) = b$.
Therefore,$\frac{dy}{dx} = \frac{b}{ab + 2ay}$.

(D) Given $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.
Squaring both sides,we get $y^2 = x + y$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$.
Rearranging the terms,$(2y - 1) \frac{dy}{dx} = 1$,which gives $\frac{dy}{dx} = \frac{1}{2y - 1}$. This matches option $A$.
From $y^2 - y = x$,we have $y(y - 1) = x$,so $y - 1 = \frac{x}{y}$.
Substituting $2y - 1 = y + (y - 1) = y + \frac{x}{y} = \frac{y^2 + x}{y}$,we get $\frac{dy}{dx} = \frac{1}{(y^2 + x)/y} = \frac{y}{y^2 + x}$. Since $y^2 = x + y$,we have $y^2 + x = (x + y) + x = 2x + y$. Thus,$\frac{dy}{dx} = \frac{y}{2x + y}$. This matches option $B$.
Solving $y^2 - y - x = 0$ for $y$ using the quadratic formula,$y = \frac{1 \pm \sqrt{1 + 4x}}{2}$. Since $y > 0$,$y = \frac{1 + \sqrt{1 + 4x}}{2}$.
Differentiating $y = \frac{1}{2} + \frac{1}{2}(1 + 4x)^{1/2}$,we get $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{2} (1 + 4x)^{-1/2} \cdot 4 = \frac{1}{\sqrt{1 + 4x}}$. This matches option $C$.
Therefore,all options are correct.

(A) Given the infinite expression $x = y^{x^{y^{x^{\dots}}}}$,we can write it as $x = y^{x^x}$.
Taking the natural logarithm on both sides,we get $\ln x = x^x \ln y$.
At $x = 1$,we have $\ln 1 = 1^1 \ln y$,which implies $0 = \ln y$,so $y = 1$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(\ln x) = \frac{d}{dx}(x^x \ln y)$
$\frac{1}{x} = \frac{d}{dx}(x^x) \cdot \ln y + x^x \cdot \frac{d}{dx}(\ln y)$
Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we have:
$\frac{1}{x} = x^x(1 + \ln x) \ln y + x^x \cdot \frac{1}{y} y'$
Substituting $x = 1$ and $y = 1$:
$\frac{1}{1} = 1^1(1 + \ln 1) \ln 1 + 1^1 \cdot \frac{1}{1} y'$
$1 = 1(1 + 0)(0) + 1 \cdot y'$
$1 = 0 + y'$
Therefore,$y' = 1$.

(A) Given the equation $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \infty}}}$.
Since the expression repeats infinitely,we can write it as $y = \sqrt{\sin x + y}$.
Squaring both sides,we get $y^2 = \sin x + y$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$.
This simplifies to $2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$.
Factoring out $\frac{dy}{dx}$,we have $\frac{dy}{dx}(2y - 1) = \cos x$.
Therefore,$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$.

(A) Given the expression $x = e^{(y+e)^{(y+e)^{(y+\ldots \infty)}}}$,we can observe that the exponent is a repeating structure starting from the first $(y+e)$.
Since the entire expression is equal to $x$,we can write the equation as $x = e^{y+x}$.
Taking the natural logarithm on both sides,we get $\ln(x) = y + x$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\ln(x)) = \frac{d}{dx}(y + x)$
$\frac{1}{x} = \frac{dy}{dx} + 1$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{1}{x} - 1$
$\frac{dy}{dx} = \frac{1-x}{x}$.

(A) Given the infinite continued fraction: $y = \frac{\sin x}{1 + \frac{\cos x}{1 + y}}$.
Multiplying both sides by $(1 + y + \cos x)$,we get: $y(1 + y + \cos x) = \sin x(1 + y)$.
$y + y^2 + y \cos x = \sin x + y \sin x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} + 2y \frac{dy}{dx} + (\frac{dy}{dx} \cos x - y \sin x) = \cos x(1 + y) + \sin x \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx}(1 + 2y + \cos x - \sin x) = \cos x + y \cos x + y \sin x$.
$\frac{dy}{dx} = \frac{y \sin x + (1 + y) \cos x}{1 + 2y + \cos x - \sin x}$.

(D) Given that,$ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $
Since the expression under the square root repeats infinitely,we can write:
$ y=\sqrt{x+y} $
Squaring both sides,we get:
$ y^{2}=x+y $
Differentiating both sides with respect to $ x $:
$ \frac{d}{dx}(y^{2}) = \frac{d}{dx}(x+y) $
$ 2y \frac{dy}{dx} = 1 + \frac{dy}{dx} $
Rearranging the terms to solve for $ \frac{dy}{dx} $:
$ 2y \frac{dy}{dx} - \frac{dy}{dx} = 1 $
$ \frac{dy}{dx}(2y - 1) = 1 $
$ \frac{dy}{dx} = \frac{1}{2y - 1} $

(D) Given the equation: $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$
Squaring both sides,we get: $y^2 = x + \sqrt{x + \sqrt{x + \ldots}}$
Since the expression under the square root is $y$,we can write: $y^2 = x + y$
Rearranging the terms: $y^2 - y = x$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(x)$
$(2y - 1) \frac{dy}{dx} = 1$
Therefore,$\frac{dy}{dx} = \frac{1}{2y - 1}$

(C) Given the equation $y = e^{x^2 + e^{x^2 + e^{x^2} + \dots}}$.
Since the exponent repeats infinitely,we can write $y = e^{x^2 + y}$.
Taking the natural logarithm on both sides,we get $\ln(y) = x^2 + y$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(x^2 + y)$
$\frac{1}{y} \frac{dy}{dx} = 2x + \frac{dy}{dx}$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} = 2x$
$\frac{dy}{dx} (\frac{1}{y} - 1) = 2x$
$\frac{dy}{dx} (\frac{1 - y}{y}) = 2x$
$\frac{dy}{dx} = \frac{2xy}{1 - y}$.

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

(C) Given the equation $y=\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\ldots \infty}}$.
Since the expression repeats infinitely,we can write it as $y=\sqrt{\sin (\log_{2} x)+y}$.
Squaring both sides,we get $y^2 = \sin (\log_{2} x) + y$,which simplifies to $y^2 - y = \sin (\log_{2} x)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\sin (\log_{2} x))$.
$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{d}{dx}(\log_{2} x)$.
Using the change of base formula $\log_{2} x = \frac{\ln x}{\ln 2}$,we have $\frac{d}{dx}(\log_{2} x) = \frac{1}{x \ln 2}$.
Thus,$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{1}{x \ln 2}$.
$\frac{dy}{dx} = \frac{\cos (\log_{2} x)}{x \ln 2 (2y - 1)}$.
Note: If the original expression meant $\log(2x)$,the derivative is $\frac{\cos(\log(2x))}{x(2y-1)}$. Given the options,the intended expression is $\log(2x)$.