Differentiation of Exponential and Logarithmic Questions in English

(A) Given $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}} = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} \{ \log(1+\sin x) - \log(1-\sin x) \}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left( \frac{\cos x}{1+\sin x} - \frac{-\cos x}{1-\sin x} \right) = \frac{1}{2} \cos x \left( \frac{1}{1+\sin x} + \frac{1}{1-\sin x} \right)$.
Simplifying the expression inside the bracket:
$\frac{dy}{dx} = \frac{1}{2} \cos x \left( \frac{1-\sin x + 1+\sin x}{(1+\sin x)(1-\sin x)} \right) = \frac{1}{2} \cos x \left( \frac{2}{1-\sin^2 x} \right) = \frac{1}{2} \cos x \left( \frac{2}{\cos^2 x} \right) = \frac{1}{\cos x} = \sec x$.
At $x = \frac{\pi}{3}$,$\frac{dy}{dx} = \sec \left( \frac{\pi}{3} \right) = 2$.

(D) Let $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}}$.
Using logarithmic properties,$y = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} [\log(1+\sin x) - \log(1-\sin x)]$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1+\sin x} \cdot \cos x - \frac{1}{1-\sin x} \cdot (-\cos x) \right]$
$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{\cos x}{1+\sin x} + \frac{\cos x}{1-\sin x} \right]$
$\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{1-\sin x + 1+\sin x}{(1+\sin x)(1-\sin x)} \right]$
$\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{2}{1-\sin^2 x} \right] = \frac{\cos x}{\cos^2 x} = \sec x$.

(B) Given $y=2^{ax}$.
Applying the derivative formula $\frac{d}{dx}(b^{f(x)}) = b^{f(x)} \cdot \ln(b) \cdot f'(x)$,we get:
$\frac{dy}{dx} = 2^{ax} \cdot \ln(2) \cdot a$.
At $x=1$,the derivative is $\left(\frac{dy}{dx}\right)_{x=1} = 2^a \cdot a \cdot \ln(2)$.
Given that $\left(\frac{dy}{dx}\right)_{x=1} = \log 256$,and knowing $\log 256 = \log(2^8) = 8 \log 2$,we equate the two expressions:
$2^a \cdot a \cdot \ln(2) = 8 \ln(2)$.
Dividing both sides by $\ln(2)$,we get:
$a \cdot 2^a = 8$.
By inspection,if $a=2$,then $2 \cdot 2^2 = 2 \cdot 4 = 8$.
Thus,$a=2$.

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.
Rationalizing the denominator inside the logarithm:
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})(x + \sqrt{x^2 + 25})}{(\sqrt{x^2 + 25} - x)(\sqrt{x^2 + 25} + x)} \right]$
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{(x^2 + 25) - x^2} \right]$
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{25} \right]$
Using logarithmic properties:
$y = 2 \log (x + \sqrt{x^2 + 25}) - \log 25$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{x + \sqrt{x^2 + 25}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 25}) - 0$
$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( 1 + \frac{1}{2\sqrt{x^2 + 25}} \cdot 2x \right)$
$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( \frac{\sqrt{x^2 + 25} + x}{\sqrt{x^2 + 25}} \right)$
$\frac{dy}{dx} = \frac{2}{\sqrt{x^2 + 25}}$

(B) Let $y = \log _{e^2}(\log x)$.
Using the logarithmic property $\log _{b^n} a = \frac{1}{n} \log _b a$,we can rewrite the expression as:
$y = \frac{1}{2} \log _e(\log x)$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \frac{1}{2} \log _e(\log x) \right]$
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\log x} \cdot \frac{1}{x}$
$\frac{dy}{dx} = \frac{1}{2x \log x}$
Thus,the correct option is $B$.

(A) Given,$y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10$.
Using the change of base formula $\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we can rewrite the expression as:
$y = \frac{\log_{e} x}{\log_{e} 10} + \frac{\log_{e} 10}{\log_{e} x} + 1 + 1$.
$y = \frac{1}{\log_{e} 10} \cdot \log_{e} x + \log_{e} 10 \cdot (\log_{e} x)^{-1} + 2$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_{e} 10} \cdot \frac{d}{dx}(\log_{e} x) + \log_{e} 10 \cdot \frac{d}{dx}((\log_{e} x)^{-1}) + 0$.
$\frac{dy}{dx} = \frac{1}{\log_{e} 10} \cdot \frac{1}{x} + \log_{e} 10 \cdot (-1)(\log_{e} x)^{-2} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} - \frac{\log_{e} 10}{x(\log_{e} x)^{2}}$.

(A) Given,$y = \log_{\cos x} \sin x = \frac{\log \sin x}{\log \cos x}$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \log \sin x$ and $v = \log \cos x$.
Then $u' = \frac{1}{\sin x} \cdot \cos x = \cot x$ and $v' = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
$\frac{dy}{dx} = \frac{(\log \cos x)(\cot x) - (\log \sin x)(-\tan x)}{(\log \cos x)^2}$.
$\frac{dy}{dx} = \frac{\cot x \log \cos x + \tan x \log \sin x}{(\log \cos x)^2}$.

(C) We have,$y = \log |x| = \begin{cases} \log x, & x > 0 \\ \log (-x), & x < 0 \end{cases}$
For $x > 0$,$\frac{dy}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$.
For $x < 0$,$\frac{dy}{dx} = \frac{d}{dx}(\log (-x)) = \frac{1}{-x} \times \frac{d}{dx}(-x) = \frac{1}{-x} \times (-1) = \frac{1}{x}$.
Combining both cases,we get $\frac{dy}{dx} = \frac{1}{x}$ for all $x \neq 0$.

(A) Given $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$.
Let $u = x^x$. Then $f(x) = \cot^{-1}\left(\frac{u - u^{-1}}{2}\right) = \cot^{-1}\left(\frac{u^2 - 1}{2u}\right)$.
Using the identity $\cot^{-1}(z) = \tan^{-1}(1/z)$,we have $f(x) = \tan^{-1}\left(\frac{2u}{u^2 - 1}\right)$.
Recall that $2\tan^{-1}(u) = \tan^{-1}\left(\frac{2u}{1 - u^2}\right)$,so $f(x) = -2\tan^{-1}(u) + \frac{\pi}{2}$ (for appropriate range).
Differentiating with respect to $x$: $f'(x) = -2 \cdot \frac{1}{1 + u^2} \cdot \frac{du}{dx}$.
Since $u = x^x$,$\frac{du}{dx} = x^x(1 + \ln x)$.
At $x = 1$,$u = 1^1 = 1$ and $\frac{du}{dx} = 1^1(1 + \ln 1) = 1$.
Thus,$f'(1) = -2 \cdot \frac{1}{1 + 1^2} \cdot 1 = -2 \cdot \frac{1}{2} = -1$.

(C) The formula for continuous compounding is $A = P \cdot e^{rt}$,where $P$ is the principal amount,$r$ is the annual interest rate,and $t$ is the time in years.
Given: $P = 10000$,$r = 4 \% = 0.04$,and $t = 10$ years.
Substituting the values: $A = 10000 \times e^{(0.04 \times 10)} = 10000 \times e^{0.4}$.
Using the given value $e^{0.4} = 1.49182$:
$A = 10000 \times 1.49182 = 14918.2$.
Rounding to the nearest integer,the amount is $Rs. 14918$.

(A) Let $y = e^{x \log x} + e^3$.
Since $e^{x \log x} = e^{\log(x^x)} = x^x$,the expression becomes $y = x^x + e^3$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(e^3)$.
Since $e^3$ is a constant,its derivative is $0$.
To differentiate $x^x$,let $u = x^x$. Then $\log u = x \log x$.
Differentiating both sides with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x)$.
Therefore,$\frac{dy}{dx} = x^x(1 + \log x) + 0 = x^x(1 + \log x)$.

(A) Let $f(x) = \log \left(\frac{1}{x}\right) + \log \left(\frac{1}{x^2}\right) + \log \left(\frac{1}{x^3}\right)$.
Using the property $\log(a^b) = b \log(a)$ and $\log(\frac{1}{x}) = -\log(x)$,we can simplify the expression:
$f(x) = -\log(x) - 2\log(x) - 3\log(x)$
$f(x) = -6\log(x)$.
Now,differentiate with respect to $x$:
$\frac{d}{dx} f(x) = \frac{d}{dx} (-6\log(x))$
$= -6 \times \frac{1}{x} = -\frac{6}{x}$.
Thus,the correct option is $A$.

(A) We know that $\log _{|x|} e = \frac{1}{\log _e |x|} = \frac{1}{\ln |x|}$.
Let $y = \frac{1}{\ln |x|}$.
Using the chain rule,$\frac{d y}{d x} = \frac{d}{d x}(\ln |x|)^{-1} = -1 \cdot (\ln |x|)^{-2} \cdot \frac{d}{d x}(\ln |x|)$.
Since $\frac{d}{d x}(\ln |x|) = \frac{1}{x}$,we have $\frac{d y}{d x} = -(\ln |x|)^{-2} \cdot \frac{1}{x} = \frac{-1}{x(\ln |x|)^2}$.
Thus,the correct option is $A$.

(B) We know that $\log_a b = \frac{\log_e b}{\log_e a}$.
So,$\log_5 x^2 = \frac{\ln x^2}{\ln 5} = \frac{2 \ln x}{\ln 5}$.
Now,differentiating with respect to $x$:
$\frac{d}{dx} \left( \frac{2 \ln x}{\ln 5} \right) = \frac{2}{\ln 5} \cdot \frac{d}{dx}(\ln x)$.
Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$,we get:
$\frac{2}{\ln 5} \cdot \frac{1}{x} = \frac{2}{x \ln 5}$.
Since $\ln 5 = \log 5$,the result is $\frac{2}{x \log 5}$.

(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$
We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for $|x| < 1$.
Therefore,$y = e^{\log _{e} (1-x)^{-1}}$.
Using the property $e^{\log _{e} f(x)} = f(x)$,we get $y = (1-x)^{-1}$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x} = \frac{d}{d x} (1-x)^{-1} = -1(1-x)^{-2} \cdot \frac{d}{d x}(1-x)$
$\frac{d y}{d x} = -1(1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \frac{1}{(1-x)^{2}}$.

(A) Given $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$.
Using the chain rule and the quotient rule for differentiation:
$\frac{dy}{dx} = \frac{1}{\frac{1-x^{2}}{1+x^{2}}} \cdot \frac{d}{dx} \left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\frac{dy}{dx} = \frac{1+x^{2}}{1-x^{2}} \cdot \left[ \frac{(-2x)(1+x^{2}) - (2x)(1-x^{2})}{(1+x^{2})^{2}} \right]$
$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-2x - 2x^{3} - 2x + 2x^{3}}{1+x^{2}} \right]$
$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-4x}{1+x^{2}} \right]$
$\frac{dy}{dx} = \frac{-4x}{(1-x^{2})(1+x^{2})}$
$\frac{dy}{dx} = \frac{-4x}{1-x^{4}}$

(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.
Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e}(\log _{e} x)}{\log _{e} x}$.
Applying the quotient rule,$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot \frac{d}{dx}(\log _{e}(\log _{e} x)) - \log _{e}(\log _{e} x) \cdot \frac{d}{dx}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot (\frac{1}{\log _{e} x} \cdot \frac{1}{x}) - \log _{e}(\log _{e} x) \cdot \frac{1}{x}}{(\log _{e} x)^{2}} \right]$.
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{1}{x} \log _{e}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.
At $x = e$,$\log _{e} e = 1$ and $\log _{e}(\log _{e} e) = \log _{e} 1 = 0$.
Substituting these values,$f^{\prime}(e) = \frac{1}{2} \left[ \frac{\frac{1}{e} - \frac{1}{e} \cdot 0}{(1)^{2}} \right] = \frac{1}{2e}$.

(D) Given,$y=2^{\log x}$.
Applying the chain rule and the derivative formula $\frac{d}{dx}(a^u) = a^u \cdot \ln a \cdot \frac{du}{dx}$,where $u = \log x$ and $a = 2$.
$\frac{dy}{dx} = \frac{d}{dx}(2^{\log x}) = 2^{\log x} \cdot \ln 2 \cdot \frac{d}{dx}(\log x)$.
Since $\frac{d}{dx}(\log x) = \frac{1}{x}$,we have:
$\frac{dy}{dx} = 2^{\log x} \cdot \ln 2 \cdot \frac{1}{x}$.
Therefore,$\frac{dy}{dx} = \frac{2^{\log x} \cdot \log 2}{x}$.

(A) For Statement $1$:
$y = \log_{10} x + \log_{e} x$
Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:
$y = \frac{\log_{e} x}{\log_{e} 10} + \log_{e} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} + \frac{1}{x}$
Since $\frac{1}{\log_{e} 10} = \log_{10} e$,we get:
$\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.
Thus,Statement $1$ is true.
For Statement $2$:
$\frac{d}{dx}(\log_{10} x) = \frac{d}{dx} \left( \frac{\log_{e} x}{\log_{e} 10} \right) = \frac{1}{x \log_{e} 10}$.
The statement claims it is $\frac{\log x}{\log 10}$,which is incorrect.
Similarly,$\frac{d}{dx}(\log_{e} x) = \frac{1}{x}$.
The statement claims it is $\frac{\log x}{\log e}$,which is incorrect.
Thus,Statement $2$ is false.

(B) Given the function $f(x) = x^{x}$.
Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.
Differentiating both sides with respect to $x$ using the chain rule and product rule:
$\frac{1}{f(x)} f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$f'(x) = f(x)(\ln x + 1) = x^{x}(\ln x + 1)$.
$A$ stationary point occurs where $f'(x) = 0$.
Since $x^{x} > 0$ for $x > 0$,we set $\ln x + 1 = 0$.
$\ln x = -1$.
Converting from logarithmic to exponential form,$x = e^{-1} = \frac{1}{e}$.

(A) Given curves are $y=a^x$ and $y=b^x$.
At the point of intersection,$a^x = b^x$,which implies $x=0$.
Thus,the curves intersect at the point $(0, 1)$.
Let $m_1$ be the slope of the tangent to $y=a^x$ at $x=0$.
$m_1 = \left. \frac{dy}{dx} \right|_{x=0} = \left. a^x \ln a \right|_{x=0} = \ln a$.
Let $m_2$ be the slope of the tangent to $y=b^x$ at $x=0$.
$m_2 = \left. \frac{dy}{dx} \right|_{x=0} = \left. b^x \ln b \right|_{x=0} = \ln b$.
The angle $\alpha$ between the curves is given by $\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,we get $\tan \alpha = \frac{\ln a - \ln b}{1 + \ln a \ln b}$.

(A) Let $y = \log \left\{e^x \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}$.
Using logarithmic properties,we have:
$y = \log e^x + \log \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}} = x \log e + \frac{3}{4} \log \left(\frac{x-2}{x+2}\right) = x + \frac{3}{4} [\log(x-2) - \log(x+2)]$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x-2} - \frac{1}{x+2} \right] = 1 + \frac{3}{4} \left[ \frac{(x+2) - (x-2)}{(x-2)(x+2)} \right] = 1 + \frac{3}{4} \left[ \frac{4}{x^2-4} \right] = 1 + \frac{3}{x^2-4}$.
At $x = 3$:
$\left(\frac{dy}{dx}\right)_{x=3} = 1 + \frac{3}{3^2-4} = 1 + \frac{3}{9-4} = 1 + \frac{3}{5} = \frac{8}{5}$.

(B) Given $x < 0$,we have $|x| = -x$.
Substituting this into the expression,we get $y = (-x)^x$.
Taking the natural logarithm on both sides: $\log y = x \log(-x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x] \cdot \log(-x) + x \cdot \frac{d}{dx} [\log(-x)]$.
$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log(-x) + x \cdot \frac{1}{-x} \cdot (-1) = \log(-x) + 1$.
Therefore,$\frac{dy}{dx} = y [\log(-x) + 1] = (-x)^x [1 + \log(-x)]$.

(D) $f(x) = \log _{x^2}(\log x)$
Using the change of base formula $\log _a b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(\log x)}{\log(x^2)} = \frac{\log(\log x)}{2 \log x}$
Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{1}{2} \frac{d}{dx} \left( \frac{\log(\log x)}{\log x} \right)$
$f'(x) = \frac{1}{2} \left[ \frac{(\log x) \cdot \frac{d}{dx}(\log(\log x)) - \log(\log x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \right]$
$f'(x) = \frac{1}{2} \left[ \frac{(\log x) \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \log(\log x) \cdot \frac{1}{x}}{(\log x)^2} \right]$
$f'(x) = \frac{1}{2x} \left[ \frac{1 - \log(\log x)}{(\log x)^2} \right]$
At $x = e$,$\log x = \log e = 1$ and $\log(\log x) = \log(1) = 0$:
$f'(e) = \frac{1}{2e} \left[ \frac{1 - 0}{(1)^2} \right] = \frac{1}{2e} = (2e)^{-1}$

(C) Let $y = \log \left( \sqrt{x + \sqrt{x^2 + a^2}} \right)$.
Using the property of logarithms,$\log(u^n) = n \log(u)$,we can simplify the expression:
$y = \frac{1}{2} \log \left( x + \sqrt{x^2 + a^2} \right)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right)$.
Calculate the derivative of the inner term:
$\frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right) = 1 + \frac{1}{2 \sqrt{x^2 + a^2}} \cdot (2x) = 1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}$.
Substitute this back into the derivative expression:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left( \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \right)$.
Cancel the common term $(x + \sqrt{x^2 + a^2})$:
$\frac{dy}{dx} = \frac{1}{2 \sqrt{x^2 + a^2}}$.

(D) Given $y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right)$.
Rationalizing the denominator inside the logarithm:
$y = \log \left( \frac{(\sqrt{x^2+1}-x)^2}{(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)} \right) = \log \left( \frac{(\sqrt{x^2+1}-x)^2}{(x^2+1)-x^2} \right) = \log (\sqrt{x^2+1}-x)^2$.
Using the property $\log(a^b) = b \log a$,we get $y = 2 \log(\sqrt{x^2+1}-x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{x^2+1}-x} \cdot \left( \frac{1}{2\sqrt{x^2+1}} \cdot 2x - 1 \right)$.
$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{x^2+1}-x} \cdot \left( \frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}} \right)$.
$\frac{dy}{dx} = 2 \cdot \frac{-( \sqrt{x^2+1}-x )}{\sqrt{x^2+1}(\sqrt{x^2+1}-x)} = \frac{-2}{\sqrt{x^2+1}}$.

(C) Given $y = \log_2(\log_2 x)$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:
$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2}$.
$y = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$.
Since $\log_e(\log_e 2)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$.
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x \log_e x \log_e 2}$.

(A) Given $f(x) = \cot^{-1} \left(\frac{x^{x} - x^{-x}}{2}\right)$.
Applying the chain rule,$f'(x) = -\frac{1}{1 + \left(\frac{x^x - x^{-x}}{2}\right)^2} \cdot \frac{d}{dx} \left(\frac{x^x - x^{-x}}{2}\right)$.
Simplifying the expression,$f'(x) = -\frac{4}{4 + (x^x - x^{-x})^2} \cdot \frac{1}{2} \cdot \frac{d}{dx} (x^x - x^{-x})$.
Note that $(x^x + x^{-x})^2 = (x^x - x^{-x})^2 + 4$,so $f'(x) = -\frac{2}{(x^x + x^{-x})^2} \cdot \frac{d}{dx} (x^x - x^{-x})$.
Using $\frac{d}{dx}(x^x) = x^x(1 + \log x)$ and $\frac{d}{dx}(x^{-x}) = -x^{-x}(1 + \log x)$,we get $\frac{d}{dx}(x^x - x^{-x}) = (x^x + x^{-x})(1 + \log x)$.
Thus,$f'(x) = -\frac{2}{(x^x + x^{-x})^2} \cdot (x^x + x^{-x})(1 + \log x) = -\frac{2(1 + \log x)}{x^x + x^{-x}}$.
At $x = 1$,$f'(1) = -\frac{2(1 + \log 1)}{1^1 + 1^{-1}} = -\frac{2(1 + 0)}{1 + 1} = -\frac{2}{2} = -1$.

(B) Given,$y = \log_2(\log_2 x)$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:
$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2} = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$.
Since $\log_e(\log_e 2)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.
Therefore,$\frac{dy}{dx} = \frac{1}{x \log_e x \log_e 2}$.

(B) Let $I = \int \frac{1}{x \sqrt{1-x^3}} dx$. Multiply numerator and denominator by $x^2$:
$I = \int \frac{x^2}{x^3 \sqrt{1-x^3}} dx$.
Let $t = \sqrt{1-x^3}$,then $t^2 = 1-x^3$,so $x^3 = 1-t^2$.
Differentiating,$2t dt = -3x^2 dx$,so $x^2 dx = -\frac{2}{3} t dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{2}{3} t dt}{(1-t^2) t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = \frac{2}{3} \int \frac{dt}{t^2-1} = \frac{2}{3} \cdot \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| + C = \frac{1}{3} \log \left| \frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1} \right| + C$.
Comparing this with the given expression $A \log \left( \frac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1} \right)$,we get $A = \frac{1}{3}$ and $B = -1$.
Therefore,$AB = \frac{1}{3} \times (-1) = -\frac{1}{3}$.

(A) Given the curve equation: $e^{y} = 1 + x^2$.
Taking the natural logarithm on both sides: $y = \ln(1 + x^2)$.
To find the slope $m$,we differentiate $y$ with respect to $x$:
$m = \frac{dy}{dx} = \frac{d}{dx}(\ln(1 + x^2)) = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2)$.
$m = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2}$.
Now,evaluate the slope at $x = 1$:
$m = \frac{2(1)}{1 + (1)^2} = \frac{2}{1 + 1} = \frac{2}{2} = 1$.
Thus,$m = 1$.

(D) Given $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$.
We can rewrite the base and the argument as:
$x^2-2x+1 = (x-1)^2$
$x^2-3x+2 = (x-1)(x-2)$
So,$f(x) = \log_{(x-1)^2} ((x-1)(x-2))$.
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we get:
$f(x) = \frac{1}{2} \log_{(x-1)} ((x-1)(x-2)) = \frac{1}{2} [\log_{(x-1)} (x-1) + \log_{(x-1)} (x-2)]$
$f(x) = \frac{1}{2} [1 + \log_{(x-1)} (x-2)] = \frac{1}{2} [1 + \frac{\ln(x-2)}{\ln(x-1)}]$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{1}{2} \left[ \frac{\ln(x-1) \cdot \frac{1}{x-2} - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right]$.
Substitute $x = 3$:
$f'(3) = \frac{1}{2} \left[ \frac{\ln(2) \cdot \frac{1}{1} - \ln(1) \cdot \frac{1}{2}}{(\ln(2))^2} \right] = \frac{1}{2} \left[ \frac{\ln(2)}{(\ln(2))^2} \right] = \frac{1}{2 \ln(2)} = \frac{1}{\ln(4)} = \log_4 e$.

(D) Given,$f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$
Using the property $\log(ab) = \log a + \log b$,we get:
$f(x)=\log _e e^{2 x}+\log _e\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}$
$f(x)=2 x+\frac{2}{3} \log _e\left(\frac{3 x+5}{5-3 x}\right)$
Using $\log(\frac{a}{b}) = \log a - \log b$:
$f(x)=2 x+\frac{2}{3}\left(\log _e(3 x+5)-\log _e(5-3 x)\right)$
Differentiating with respect to $x$:
$\frac{d f}{d x}=2+\frac{2}{3}\left(\frac{3}{3 x+5}-\frac{-3}{5-3 x}\right)$
$\frac{d f}{d x}=2+2\left(\frac{1}{3 x+5}+\frac{1}{5-3 x}\right)$
At $x=1$:
$\left(\frac{d f}{d x}\right)_{x=1}=2+2\left(\frac{1}{3(1)+5}+\frac{1}{5-3(1)}\right)$
$=2+2\left(\frac{1}{8}+\frac{1}{2}\right)=2+2\left(\frac{1+4}{8}\right)=2+2\left(\frac{5}{8}\right)=2+\frac{5}{4}=\frac{13}{4}$

(C) Let $f(x) = (\log x)^{\sin x}$ and $g(x) = \cos x$.
We need to find $\frac{df}{dg} = \frac{f'(x)}{g'(x)}$.
Taking the natural logarithm of $f(x)$: $\log(f(x)) = \sin x \cdot \log(\log x)$.
Differentiating with respect to $x$:
$\frac{1}{f(x)} f'(x) = \cos x \cdot \log(\log x) + \sin x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$.
Thus,$f'(x) = (\log x)^{\sin x} \left[ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right]$.
Also,$g'(x) = -\sin x$.
Therefore,$\frac{df}{dg} = \frac{(\log x)^{\sin x} [ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} ]}{-\sin x}$.
At $x = \frac{\pi}{2}$,$\sin x = 1$,$\cos x = 0$,and $\log x = \log(\frac{\pi}{2})$.
Substituting these values:
$\frac{df}{dg} = \frac{(\log(\frac{\pi}{2}))^1 [ 0 \cdot \log(\log(\frac{\pi}{2})) + \frac{1}{(\frac{\pi}{2}) \log(\frac{\pi}{2})} ]}{-1}$.
$= - \frac{1}{\frac{\pi}{2}} = -\frac{2}{\pi}$.

(A) Let $u = (\sin x)^x$ and $v = x^{\sin x}$.
Taking log on both sides for $u$:
$\log u = x \log (\sin x)$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log (\sin x) + x \cot x$.
$\frac{du}{dx} = (\sin x)^x [\log (\sin x) + x \cot x] = (\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]$.
Taking log on both sides for $v$:
$\log v = \sin x \log x$.
Differentiating with respect to $x$:
$\frac{1}{v} \frac{dv}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{x \cos x \log x + \sin x}{x}$.
$\frac{dv}{dx} = x^{\sin x} \cdot \frac{x \cos x \log x + \sin x}{x} = x^{\sin x-1} [\sin x + x \cos x \log x]$.
Therefore,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]}{x^{\sin x-1} [\sin x + x \cos x \log x]}$.

(B) Given that $f(x)=e^x$ and $g(x)=\sin^{-1} x$.
Then $h(x) = f(g(x)) = e^{\sin^{-1} x}$.
Taking the natural logarithm on both sides,we get $\ln(h(x)) = \sin^{-1} x$.
Differentiating both sides with respect to $x$,we get $\frac{1}{h(x)} \cdot h'(x) = \frac{d}{dx}(\sin^{-1} x)$.
Thus,$\frac{h'(x)}{h(x)} = \frac{1}{\sqrt{1-x^2}}$.

(A) Given the curve $y = e^{a+bx^2}$.
First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{d}{dx}(e^{a+bx^2}) = e^{a+bx^2} \cdot \frac{d}{dx}(a+bx^2) = e^{a+bx^2} \cdot (2bx) = 2bxy$.
At the point $P(1,1)$,the slope is $-2$:
$\left. \frac{dy}{dx} \right|_{(1,1)} = 2b(1)(1) = -2$.
$2b = -2 \implies b = -1$.
Since the point $P(1,1)$ lies on the curve,it must satisfy the equation:
$1 = e^{a+b(1)^2} \implies 1 = e^{a+b}$.
Taking the natural logarithm on both sides:
$\ln(1) = a+b \implies 0 = a+b$.
Substituting $b = -1$:
$0 = a - 1 \implies a = 1$.
Finally,calculate $2a - 3b$:
$2a - 3b = 2(1) - 3(-1) = 2 + 3 = 5$.

(C) Let $y_n = \log^n x$. Then $y_n = \log(y_{n-1})$,where $y_1 = \log x$,$y_2 = \log(\log x)$,and so on.
By the chain rule,$\frac{dy_n}{dx} = \frac{d}{dx}(\log y_{n-1}) = \frac{1}{y_{n-1}} \cdot \frac{dy_{n-1}}{dx}$.
Expanding this,we get $\frac{dy_n}{dx} = \frac{1}{y_{n-1}} \cdot \frac{1}{y_{n-2}} \cdot \dots \cdot \frac{1}{y_1} \cdot \frac{d}{dx}(\log x)$.
Since $\frac{d}{dx}(\log x) = \frac{1}{x}$,we have $\frac{dy_n}{dx} = \frac{1}{y_{n-1} y_{n-2} \dots y_1 x}$.
Substituting $y_k = \log^k x$,we get $\frac{dy}{dx} = \frac{1}{x \log x \log^2 x \dots \log^{n-1} x}$.
Therefore,$x \log x \log^2 x \dots \log^{n-1} x \frac{dy}{dx} = 1$.

(C) Given $f(x) = \log_{5} \log_{3} x$.
Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$,we have:
$f(x) = \frac{\ln(\log_{3} x)}{\ln 5} = \frac{\ln(\frac{\ln x}{\ln 3})}{\ln 5} = \frac{\ln(\ln x) - \ln(\ln 3)}{\ln 5}$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx} [\ln(\ln x) - \ln(\ln 3)] = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$.
Now,substituting $x = e$:
$f^{\prime}(e) = \frac{1}{\ln 5} \cdot \frac{1}{\ln e} \cdot \frac{1}{e} = \frac{1}{\ln 5 \cdot 1 \cdot e} = \frac{1}{e \ln 5}$.
Since $\ln 5 = \log_{e} 5$,we get $f^{\prime}(e) = \frac{1}{e \log_{e} 5}$.

(A) Let $y = 5^{\log x}$.
Using the property $a^{\log_b c} = c^{\log_b a}$,we can write $y = x^{\log 5}$.
Now,differentiate with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$\frac{dy}{dx} = (\log 5) x^{\log 5 - 1}$.
Alternatively,using the chain rule for $y = 5^{\log x}$:
$\frac{dy}{dx} = 5^{\log x} \cdot \ln 5 \cdot \frac{d}{dx}(\log x) = 5^{\log x} \cdot \ln 5 \cdot \frac{1}{x} = \frac{5^{\log x} \ln 5}{x}$.
Since $\ln 5$ is equivalent to $\log_e 5$,the correct expression is $\frac{5^{\log x} \ln 5}{x}$.

(C) Let $u = \log_{2025} x = \frac{\log x}{\log 2025}$.
Then $y = \log_{2026} u = \frac{\log u}{\log 2026}$.
Applying the chain rule:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u \log 2026} \cdot \frac{d}{dx}\left(\frac{\log x}{\log 2025}\right)$.
$\frac{dy}{dx} = \frac{1}{u \log 2026} \cdot \frac{1}{x \log 2025}$.
Substituting $u = \frac{\log x}{\log 2025}$:
$\frac{dy}{dx} = \frac{1}{(\frac{\log x}{\log 2025}) \log 2026 \cdot x \log 2025} = \frac{1}{x \log x \log 2026}$.