Differentiation

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{d x}{d t}=1-\cos t \\ & \frac{d y}{d t}=\sin t \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\sin t}{1-\cos t}=\frac{2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \sin ^2\left(\frac{t}{2}\right)} \\ & \Rightarrow \quad \frac{d y}{d x}=\cot \left(\frac{t}{2}\right) \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-\operatorname{cosec}^2\left(\frac{t}{2}\right) \times \frac{1}{2} \times \frac{d t}{d x} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{d^2 y}{d x^2}=-\frac{1}{2 \sin ^2\left(\frac{t}{2}\right)} \times \frac{1}{1-\cos t} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{-1}{(1-\cos t)^2} \\ & \Rightarrow \quad \frac{-1}{(1-\cos k)^2}=-1 \\ & \Rightarrow \quad(1-\cos k)^2=1 \\ & \Rightarrow \quad 1-\cos k= \pm 1 \\ & \Rightarrow \quad \cos k=0 \Rightarrow k=\frac{\pi}{2} \\ & \Rightarrow\quad \lim _{t \rightarrow \frac{\pi}{2}} \frac{1-\cos t}{t-\sin t}=\frac{1-0}{\frac{\pi}{2}-1}=\frac{2}{\pi-2} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=\tan ^2\left(\cos ^{-1} \sqrt{\frac{1+x^2}{2}}\right) \\ & \text { Put } \frac{1+x^2}{2}=\cos ^2(2 \theta) \\ & \qquad x^2=2 \cos ^2(2 \theta)-1=\cos 4 \theta \\ & \text { Also, } 0 \leq \frac{1+x^2}{2} \leq 1 \\ & \quad 0 \leq 1+x^2 \leq 2 \\ & \Rightarrow \quad x^2 \leq 1 \Rightarrow-1 \leq x \leq 1 \\ & \Rightarrow \quad-1 \leq \cos 4 \theta \leq 1 \end{aligned} \end{aligned} $

$ \Rightarrow \quad 0 \leq 4 \theta \leq \pi \Rightarrow 0 \leq \theta \leq \frac{\pi}{4} $

Differentiating with resped to $x$

$ \begin{array}{ll} \Rightarrow & 2 x \frac{d x}{d \theta}=-4 \sin 4 \theta \\ \Rightarrow & \frac{d x}{d \theta}=\frac{-2 \sin 4 \theta}{\sqrt{\cos 4 \theta}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

and $\quad y=\tan ^2 \cos ^{-1}(\cos 2 \theta)=\tan ^2(2\theta)$

$ \begin{aligned} \Rightarrow \quad \frac{d y}{d \theta} & =2 \tan (2 \theta) \times \sec ^2(2 \theta) \times 2 \ldots \text { (i) } \\ \Rightarrow \quad \frac{d y}{d \theta} & =\frac{4 \sin (2 \theta)}{\cos ^3(2 \theta)} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{4 \sin (2 \theta) \times \sqrt{\cos 4 \theta}}{\cos ^3(2 \theta)(-4) \sin 2 \theta \cdot \cos 2 \theta} \\ & =-\frac{\sqrt{\cos 4 \theta}}{\cos ^4(2 \theta)} \\ & =\frac{-\sqrt{x^2}}{\left(\frac{1+\cos 4 \theta}{2}\right)^2}=\frac{-4 x}{\left(1+x^2\right)^2} \end{aligned} $

$ \begin{aligned} y= & (x)^{\ln x}+(\ln x)^x \\ y= & e^{\ln x \ln x}+e^{x \ln (\ln x)} \\ \frac{d y}{d x}= & (x)^{\ln x}\left(\frac{2 \ln x}{x}\right) +(\ln x)^x\left(\ln (\ln x)+\frac{x}{\ln x} \times \frac{1}{x}\right) \end{aligned} $

$ \begin{aligned} \left(\frac{d y}{d x}\right)_{\text {at } x=0} & =(e)\left(\frac{2}{e}\right)+1^e\left(0+\frac{e}{1} \times \frac{1}{e}\right) \\ & =2+1=3 \end{aligned} $

$ \begin{aligned} &\begin{aligned} y & =\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)}}} \ldots \infty \\ & \Rightarrow y=\sqrt{\log \left(x^2+1\right)+y} \\ & \Rightarrow y^2=\log \left(x^2+1\right)+y \\ & \Rightarrow y^2-y=\log \left(x^2+1\right) \end{aligned}\\ &\text { Differentiating wrt } x \text {, we get }\\ &\begin{aligned} & (2 y-1) \frac{d y}{d x}=\frac{2 x}{x^2+1} \\ & \frac{d y}{d x}=\frac{2 x}{(2 y-1)\left(x^2+1\right)} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } x=\sqrt{1-\tan y} \\ & \tan y=1-x^2 \end{aligned} $

Differentiating w.r.t $x$, we get

$ \begin{gathered} \sec ^2 y \frac{d y}{d x}=-2 x \\ \Rightarrow \quad \frac{d y}{d x}=\frac{-2 x}{\sec ^2 y}=\frac{-2 x}{1+\left(1-x^2\right)^2} \\ =\frac{-2 x}{x^4-2 x^2+2} \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} x & =\sin 2 \theta \cos 3 \theta, y=\sin 3 \theta \cos 2 \theta \\ \Rightarrow 2 x & =2 \sin 2 \theta \cos 3 \theta, 2 y=2 \sin 3 \theta \cdot \cos 2 \theta \\ \Rightarrow 2 x & =\sin 5 \theta-\sin \theta, 2 y=\sin 5 \theta+\sin \theta \\ \frac{d y}{d x} & =\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{5 \cos 5 \theta+\cos \theta}{5 \cos 5 \theta-\cos \theta} \\ & =\frac{4 \cos 5 \theta+\cos 5 \theta+\cos \theta}{4 \cos 5 \theta+\cos 5 \theta-\cos \theta} \\ & =\frac{2 \cos 5 \theta+\cos 3 \theta \cdot \cos 2 \theta}{2 \cos 5 \theta-\sin 3 \theta \cdot \sin 2 \theta} \end{aligned} \end{aligned} $

$ \text { } 3^x y^x=x^{3 y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

At $x=1 \Rightarrow 3 y=1 \Rightarrow y=\frac{1}{3}$

Taking $\log$ both sides in Eq. (i), we get

$ \begin{aligned} & \Rightarrow \log 3^x+\log y^x=\log x^{3 y} \\ & \Rightarrow x \log 3+x \log y=3 y \log x \end{aligned} $

Differential w.r.t. $x$, we get

$ \log 3+\frac{x}{y} \frac{d y}{d x}+\log y=\frac{3 y}{x}+3 \log x \cdot \frac{d y}{d x} $

Put $x=1, y=\frac{1}{3}$.

$ \begin{aligned} & \log 3+3 \frac{d y}{d x}+\log \frac{1}{3}=3 \times \frac{1}{3}+3 \log 1 \frac{d y}{d x} \\ & \Rightarrow \quad 3 \frac{d y}{d x}=1 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} y & =\left(1-x^2\right) \tanh ^{-1} x \\ \frac{d y}{d x} & =\left(1-x^2\right) \cdot \frac{1}{1-x^2}+\tanh ^{-1} x(-2 x) \\ & =1-2 x \tanh ^{-1}(x) \\ \frac{d^2 y}{d x^2} & =-2 \tanh ^{-1} x-\frac{2 x}{1-x^2} \\ & =-2 \frac{\left(1-x^2\right) \tanh ^{-1} x+x}{1-x^2} \\ & =-2 \frac{(y+x)}{1-x^2} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} f(x) & =\log _{\left(x^2-2 x+1\right)} x^2-3 x+2 \\ & =\log _{(x-1)^2}(x-2)(x-1) \\ & =\frac{1}{2} \log _{(x-1)}(x-2)(x-1) \\ \Rightarrow f(x) & =\frac{1}{2}\left(1+\frac{\log (x-2)}{\log (x-1)}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow f^{\prime}(x)=\frac{1}{2}\binom{\log (x-1) \cdot \frac{1}{x-2}}{\frac{-\log (x-2) \frac{1}{(x-1)}}{(\log (x-1))^2}} \\ & \therefore f^{\prime}(3)=\frac{\frac{1}{2}\left(\log 2-\frac{\log 1}{2}\right)}{(\log 2)^2} \\ & =\frac{1}{2 \log 2}=\frac{1}{\log 4}=\log _4 e \end{aligned} $

Given, $\frac{d}{d x}\left\{\left(\frac{x-1}{x-\sqrt{x}}\right) e^{2 x+1}\right\}$

$ =\frac{x-1}{x-\sqrt{x}} e^{2 x+1} f(x) $

Here, $\frac{x-1}{x-\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$

$ \begin{gathered} =\frac{\sqrt{x}+1}{\sqrt{x}}=1+\frac{1}{\sqrt{x}} \\ \therefore \frac{d}{d x}\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}=\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1} f(x) \end{gathered} $

Now, $\frac{d}{d x}\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}$

$ \begin{aligned} & =\frac{-1}{2 x \sqrt{x}} e^{2 x+1}+\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}(2) \\ & =e^{2 x+1}\left[\frac{-1}{2 x \sqrt{x}}+2\left(1+\frac{1}{\sqrt{x}}\right)\right] \end{aligned} $

$ \begin{aligned} \therefore f(x) & =\frac{\frac{-1}{2 x \sqrt{x}}+2\left(1+\frac{1}{\sqrt{x}}\right)}{1+\frac{1}{\sqrt{x}}} \\ & =\frac{-1+4 x \sqrt{x}+4 x}{2 x \sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}+1} \\ \Rightarrow f(x) & =\frac{-1+4 x \sqrt{x}+4 x}{2 x(\sqrt{x}+1)} \\ \therefore f(4) & =\frac{-1+4(4)(2+4(4)}{2(4)(2+1)} \\ & =\frac{-1+32+16}{24}=\frac{47}{24} \end{aligned} $

Given, $y=f(\cos h x)$ and

$ f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right) $

Let $u=\cos h x$, so $y=f(u)$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =f^{\prime}(u) \cdot \frac{d}{d x}(\cos h x) \\ & =f^{\prime}(\cos h x) \cdot \sin h x \end{aligned} $

Since, $f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right)$

$ \begin{aligned} \therefore f^{\prime}(\cosh x) & =\log \left(\cosh x+\sqrt{\cosh h^2 x-1}\right) \\ & =\log (\cosh x+|\sinh x|) \end{aligned} $

Since, $x>0, \sin h>0$

So, $f^{\prime}(\cos h x)=\log (\cos h x+\sin h x)$

$ \Rightarrow \log e^x=x \quad\left[\because \cosh x+\sinh x=e^x\right] $

So, $\frac{d y}{d x}=x \cdot \sin h x$

$ \begin{aligned} & \frac{d^2 y}{d x^2}=x \cdot \frac{d}{d x}(\sin h x)+\sin h x \cdot \frac{d}{d x}(x) \\ \Rightarrow & x \cdot \cos h x+\sin h x \cdot 1 \\ \Rightarrow & x \cos h x+\sin h x \end{aligned} $

Given equation

$ \left(x^2-3 x+2\right) e^{\frac{y}{x-1}}=x+2 $

Differentiate both sides w.r.t., we get

$ \begin{array}{r} \frac{d}{d x}\left(\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}}\right)=\frac{d}{d x}(x+2) \\ \Rightarrow(2 x-3) e^{\frac{y}{x-1}}+\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}} \\ \cdot \frac{d}{d x}\left(\frac{y}{x-1}\right)=1 \\ \Rightarrow(2 x-3) e^{\frac{y}{x-1}}+\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}} \\ {\left[\frac{\frac{d y}{d x}(x-1)-y}{(x-1)^2}\right]=1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Now, substitute $x=0$ into the given equation

$ \begin{array}{ll} & \left(0^2-3(0)+2\right) e^{\frac{y}{(0-1)}}=0+2 \\ \Rightarrow & 2 e^{-y}=2 \Rightarrow e^{-y}=1 \\ \Rightarrow & -y=\ln (1) \Rightarrow-y=0 \\ \Rightarrow & y=0 \end{array} $

Substitute $x=0$ and $y=0$ in Eq. (i), we get

$ \begin{aligned} & (2(0)-3) e^{\frac{0}{0-1}}+\left(0^2-3(0)+2\right) e^{\frac{0}{0-1}} \\ & \cdot\left[\frac{\frac{d y}{d x}(0-1)-0}{(0-1)^2}\right]=1 \\ & \Rightarrow-3 e^0+2 e^0\left(\frac{-d y}{d x}\right)=1 \\ & \Rightarrow \quad-3+2\left(\frac{-d y}{d x}\right)=1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad-2 \frac{d y}{d x}=1+3=4 \Rightarrow \frac{d y}{d x}=\frac{4}{-2}=-2 \\ & \therefore \quad\left(\frac{d y}{d x}\right)_{x=0}=-2 \end{aligned} $

Given $x=\frac{t^2}{1+t^3}, y=\frac{2 t^3}{1+t^5}$ and

$t \neq-1$

So, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Now, $\frac{d x}{d t}=\frac{(2 t)\left(1+t^5\right)-t^2 \cdot 5 t^4}{\left(1+t^5\right)^2}$

$ =\frac{2 t-3 t^6}{\left(1+t^5\right)^2} $

And, $\frac{d y}{d t}=\frac{\left(6 t^2\right)\left(1+t^5\right)-\left(2 t^3\right)\left(5 t^4\right)}{\left(1+t^5\right)^2}$

$ =\frac{6 t^2-4 t^7}{\left(1+t^5\right)^2} $

$ \begin{aligned}So,\,\, \frac{d y}{d x} & =\frac{\left[\frac{6 t^2-4 t^7}{\left(1+t^5\right)^2}\right]}{\left[\frac{\left(2 t-3 t^6\right)}{\left(1+t^5\right)^2}\right]}=\frac{6 t^2-4 t^7}{2 t-3 t^6} \\ & =\frac{2 t^2\left(3-2 t^5\right)}{t\left(2-3 t^5\right)}=\frac{2 t\left(3-2 t^5\right)}{\left(2-3 t^5\right)} \end{aligned} $

To solve for the derivative of the function $ f $, let's analyze the given functional equation:

$ y f(x+y) + \cos mx y = 1 + y f(x) $

Rearranging the terms, we have:

$ y[f(x+y) - f(x)] = 1 - \cos mx y $

Using the trigonometric identity for cosine, we can express this as:

$ y[f(x+y) - f(x)] = 2 \sin^2 \left(\frac{mxy}{2}\right) $

Now, by dividing both sides by $ y^2 $, we get:

$ \frac{f(x+y) - f(x)}{y} = \frac{2 \sin^2 \left(\frac{mxy}{2}\right)}{y^2} $

Taking the limit as $ y \to 0 $, this expression represents the derivative $ f'(x) $:

$ \lim\limits_{y \to 0} \frac{f(x+y) - f(x)}{y} = \lim\limits_{y \to 0} \frac{2 \sin^2 \left(\frac{mxy}{2}\right)}{y^2} $

Substitute $ m = 2 $, and notice that $\sin^2 \left(\frac{2xy}{2}\right) = \sin^2(xy)$. Therefore, the limit becomes:

$ f'(x) = \lim\limits_{y \to 0} \frac{2 \sin^2(xy)}{y^2} $

Recognizing from trigonometric limits that as $ y \to 0 $, $\sin(xy) \approx xy$, the limit simplifies through:

$ f'(x) = \lim\limits_{y \to 0} \frac{2 (xy)^2}{y^2} = \lim\limits_{y \to 0} 2x^2 = 2x^2 $

Thus, the derivative of the function is:

$ f'(x) = 2x^2 $

$ y=\sqrt{\sin (\ln 2 x)+y} $

$ y^{2}-y=\sin (\ln 2 x) $

$ (2 y-1) y^{\prime}=\cos (\ln 2 x) \times \frac{1}{2 x} \times 2 $

$ y^{\prime}=\frac{\cos (\ln 2 x)}{x(2 y-1)} $

$ y=\tan ^{-1}\left[\frac{\sin 2 x\left(\sin ^{2} 2 x-3 x^{2}\right)}{x\left(3 \sin ^{2} 2 x-x^{2}\right)}\right] $

Dividing numerator and Denominator by $ x^{3} $, we get.

$ y=\tan ^{-1}\left[\frac{\left(\frac{\sin 2 x}{x}\right)^{3}-3\left(\frac{\sin 2 x}{x}\right)}{-1+3\left(\frac{\sin 2 x}{x}\right)^{2}}\right] $

Put, $ \frac{\sin 2 x}{x}=\tan \theta $

$ \begin{array}{l} y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right]=y \\\\ =\tan ^{-1}(\tan 3 \theta)=3 \theta \\\\ y=3 \tan ^{-1}\left(\frac{\sin 2 x}{x}\right) \\\\ \frac{d y}{d x}=3 \cdot \frac{1}{1+\left(\frac{\sin 2 x}{x}\right)^{2}} \times \frac{x \cos 2 x \cdot 2-\sin 2 x}{x^{2}} \\\\ =\frac{3 x^{2}}{x^{2}+\sin ^{2} 2 x} \times \frac{2 x \cos 2 x-\sin 2 x}{x^{2}} \\\\ =\frac{6 x \cos 2 x-3 \sin 2 x}{x^{2}+\sin ^{2}(2 x)} \end{array} $

$ f=(\sin x)^{x}=e^{x \ln \sin x} $

$ \begin{array}{l} \frac{d f}{d x}=e^{x \ln \sin x}\left[x \cdot \frac{1}{\sin x} \cdot \cos x+\ln \sin x\right] \\\\ =(\sin x)^{x}[x \cot x+\ln \sin x] \\\\ g=x^{\sin x}=e^{\sin x \ln x} \\\\ \frac{d g}{d x}=e^{\sin x \cdot \ln x}\left[\sin x \cdot \frac{1}{x}+\ln x \cdot \cos x\right] \\\\ =x^{\sin x}\left[\frac{\sin x}{x}+\ln x \cdot \cos x\right] \end{array} $

$ \frac{d f}{d g}=\frac{(\sin x)^{x-1}[x \cos x+\ln \sin x \cdot(\sin x)]}{x^{\sin x-1}[\sin x+x \cos x \ln x]} $

To solve the problem, we start with the given equation:

$ y = \log(x - \sqrt{x^2 - 1}) $

We need to differentiate this with respect to $ x $. The differentiation is done as follows:

Differentiating $ y $:

$ y' = \frac{d}{dx}\left(\log(x - \sqrt{x^2 - 1})\right) = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \left(1 - \frac{x}{\sqrt{x^2 - 1}}\right) $

Simplifying the derivative:

$ y' = \frac{\sqrt{x^2 - 1} - x}{\sqrt{x^2 - 1} \cdot (x - \sqrt{x^2 - 1})} $

By simplifying further, we find:

$ y' = -\frac{1}{\sqrt{x^2 - 1}} $

Substituting back into the equation:

Multiply by $\sqrt{x^2 - 1}$, to get:

$ \sqrt{x^2 - 1} \cdot y' = -1 $

Differentiate again to find $ y'' $:

$ \frac{d}{dx}\left(\sqrt{x^2 - 1} \cdot y'\right) = 0 $

This gives:

$ \frac{2x}{2\sqrt{x^2 - 1}} \cdot y' + \sqrt{x^2 - 1} \cdot y'' = 0 $

Simplifying, we have:

$ x \cdot y' + (x^2 - 1) \cdot y'' = 0 $

Substitute in the given equation:

The required expression becomes:

$ (x^2 - 1) \cdot y'' + x \cdot y' + e^y + \sqrt{x^2 - 1} $

Since $ x \cdot y' + (x^2 - 1) \cdot y'' = 0 $, substitute $ e^y = e^{\log(x - \sqrt{x^2 - 1})} = x - \sqrt{x^2 - 1} $:

$ = 0 + (x - \sqrt{x^2 - 1}) + \sqrt{x^2 - 1} $

Thus, simplifying gives:

$ = x $

Therefore, the solution simplifies to $ x $.

Given, $ y=\log \left(\tan \sqrt{\frac{2^{x}-1}{2^{x}+1}}\right), x > 0 $

$ \frac{d y}{d x}=\frac{1}{\tan \left(\sqrt{\frac{2^{x}-1}{2^{x}+1}}\right)} $

$ \times \sec ^{2} \sqrt{\frac{2^{x}-1}{2^{x}+1}} \times\left[\frac{\sqrt{2^{x}+1}}{2 \sqrt{2^{x}-1}}\right] $

$ \times\left[\frac{2^{x} \log 2\left(2^{x}+1\right)-2^{x} \log 2\left(2^{x}-1\right)}{\left(2^{x}+1\right)^{2}}\right] $

Put $ x=1 $

$ \begin{array}{l} \frac{1}{\tan \left(\frac{1}{\sqrt{3}}\right)} \sec ^{2}\left(\frac{1}{\sqrt{3}}\right) \times \frac{1}{2} \sqrt{3} \\\\ \times\left[\frac{2 \log 2 \times 3-2 \log 2 \times 1}{9}\right] \\\\ =\frac{\sqrt{3}(4) \log 2}{2 \sin \left(\frac{1}{\sqrt{3}}\right) \cos \left(\frac{1}{\sqrt{3}}\right) \times 9}=\frac{4 \sqrt{3} \log 2}{9 \sin \left(\frac{2}{\sqrt{3}}\right)} \end{array} $

Given, $ \log y=y^{\log x} $

On taking $ \log $ both sides, we get $ \log \log y=\log x \log y $

$ \frac{1}{y \log y} \frac{d y}{d x}=\frac{\log y}{x}+\frac{\log x}{y} \frac{d y}{d x} $

$ \frac{1}{y \log y}(1-\log x \log y) \frac{d y}{d x}=\frac{\log y}{x} $

$ \frac{d y}{d x}=\frac{y(\log y)^{2}}{x(1-\log x \log y)} $

Given, $ y=a \cos 3 x+b e^{-x} $

$ \begin{aligned} y^{\prime} & =-3 a \sin 3 x-b e^{-x} \\\\ y^{\prime \prime} & =-9 a \cos 3 x+b e^{-x} \end{aligned} $

On adding Eqs. (ii) and (iii), we get

$ \begin{array}{l} \Rightarrow y^{\prime \prime}+y^{\prime}=-a[3 \sin 3 x+9 \cos 3 x] \\\\ \Rightarrow \quad a=\frac{-\left(y^{\prime \prime}+y^{\prime}\right)}{3 \sin 3 x+9 \cos 3 x} \\\\ \Rightarrow \quad \sin 3 x y^{\prime \prime}-3 \cos 3 x y^{\prime} \\\\ =(\sin 3 x+3 \cos 3 x) b e^{-x} \\\\ \Rightarrow \quad \frac{\sin 3 x y^{\prime \prime}-3 \cos 3 x y^{\prime}}{\sin 3 x+3 \cos 3 x}=b e^{-x} \\\\ \Rightarrow \quad y=\frac{-y^{\prime \prime} \cos 3 x-y^{\prime} \cos 3 x}{3(\sin 3 x+3 \cos 3 x)} \\\\ +\frac{\sin 3 x y^{\prime \prime}-3 \cos 3 x}{\sin 3 x+3 \cos 3 x} \\\\ \Rightarrow 3 y(\sin 3 x+3 \cos 3 x)=y^{\prime \prime}\left(3 \sin 3 x-\cos { }^{3}\right) \\\\ -10 \cos x x^{\prime} \\\\ \Rightarrow 10 \cos 3 x y^{\prime}+3 y(\sin 3 x+3 \cos 3 x) \\\\ =y^{\prime \prime}(3 \sin 3 x-\cos 3 x) \end{array} $

$ \begin{aligned} & \text { Given, } y=\frac{\tan x \cos ^{-1} x}{\sqrt{1-x^2}} \\\\ & \begin{aligned} & \therefore \frac{d y}{d x}=\sec ^2 x\left(\frac{\cos ^{-1} x}{\sqrt{1-x^2}}\right)+\frac{-1}{\sqrt{1-x^2}}\left(\frac{\tan x}{\sqrt{1-x^2}}\right) \\\\ &+\tan x \cos ^{-1} x\left(-\frac{1}{2}\left(1-x^2\right)^{-3 / 2}(-2 x)\right) \\\\ & \Rightarrow \frac{d y}{d x}= \frac{\sec ^2 x \cos ^{-1} x}{\sqrt{1-x^2}} \\\\ &-\frac{\tan x}{1-x^2}+\frac{\tan x \cos ^{-1} x}{\left(1-x^2\right)^{3 / 2}} \\\\ &\left.\therefore \frac{d y}{d x}\right|_{x=0}=\frac{\sec ^2 0 \cos ^{-1} 0}{\sqrt{1-0^2}} \\\\ &=\frac{\text { (1) } \frac{\pi}{2}}{1}-0+0=\frac{\pi}{2} \end{aligned} \end{aligned} $

Given, $y(\cos x)^{\sin x}=(\sin x)^{\sin x}\quad \ldots \text { (i) }$

$ \Rightarrow y=\frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}} \Rightarrow y\left(\frac{\pi}{4}\right)=\frac{\left(\frac{1}{\sqrt{2}}\right)^{\frac{1}{\sqrt{2}}}}{\left(\frac{1}{\sqrt{2}}\right)^{\frac{1}{\sqrt{2}}}}=1 $

On taking $\log$ on both sides of Eq. (i), we get

$ \begin{aligned} & \log y+(\sin x) \log (\cos x)=(\sin x) \log (\sin x) \\\\ & \Rightarrow \log y=(\sin x) \log (\tan x) \end{aligned} $

On differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} & \frac{1}{y} \frac{d y}{d x}=(\sin x) \frac{1}{\tan x} \times \sec ^2 x+\cos x \log (\tan x) \\\\ & \begin{aligned} & \therefore\left(\frac{1}{y} \frac{d y}{d x}\right)_{x=\frac{\pi}{4}} \\\\ &=\left(\sin \frac{\pi}{4}\right)^{\sec ^2\left(\frac{\pi}{4}\right)} \\\\ & \tan \left(\frac{\pi}{4}\right) \end{aligned}\left(\cos \frac{\pi}{4}\right) \log \left(\tan \frac{\pi}{4}\right) \end{aligned} $

$\begin{aligned} & \Rightarrow \frac{1}{y\left(\frac{\pi}{4}\right)} \times\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}} \\\\ & \quad=\left(\frac{1}{\sqrt{2}}\right) \frac{(\sqrt{2})^2}{(1)}+\left(\frac{1}{\sqrt{2}}\right) \log (1) \\\\ & \Rightarrow \quad \frac{1}{1} \times\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}(0) \\\\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=\sqrt{2}\end{aligned}$

Given,

$ \begin{array}{llrl} y=44 x^{45}+45 x^{-44} \\\\ \therefore y^{\prime} =(44 \times 45) x^{44}+(45 \times(-44)) x^{-45} \\\\ \therefore y^{\prime \prime} =(44 \times 45 \times 44) \\\\ x^{43}+(45 \times(-44) \times(-45)) x^{-46} \\\\ \Rightarrow y^{\prime \prime}=\frac{44 \times 45}{x^2}\left(44 x^{45}+45 x^{-44}\right) \\\\ \Rightarrow y^{\prime \prime}=\frac{1980}{x^2} y \end{array} $

We have,

$ \begin{aligned} & 2 x^2-3 x y+4 y^2+2 x-3 y+4=0 \\\\ & 4 x-\left(3 y+3 x \frac{d y}{d x}\right)+8 y \frac{d y}{d x}+2-3 \frac{d y}{d x}=0 \\\\ & \frac{d y}{d x}(-3 x+8 y-3)=3 y-4 x-2 \\\\ & \frac{d y}{d x}=\frac{3 y-4 x-2}{-3 x+8 y-3} \\\\ & \left.\frac{d y}{d x}\right|_{(3,2)} \quad=\frac{6-12-2}{-9+16-3}=\frac{-8}{4}=-2 \end{aligned} $

$ \begin{aligned} & \text { We have, } x=\frac{9 t^2}{1+t^4} \\\\ & \frac{d x}{d t}=\frac{\left(1+t^4\right) 18 t-9 t^2 \cdot 4 t^3}{\left(1+t^4\right)^2}=\frac{18 t-18 t^5}{\left(1+t^4\right)^2} \\\\ & \text { and } y=\frac{16 t^2}{1-t^4} \\\\ & \frac{d y}{d t}=\frac{\left(1-t^4\right) 32 t+16 t^2 \cdot 4 t^3}{\left(1-t^4\right)^2}=\frac{32 t+32 t^5}{\left(1-t^4\right)^2} \\\\ & \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{32 t\left(1+t^4\right)}{\left(1-t^4\right)^2}}{\frac{18 t\left(1-t^4\right)}{\left(1+t^4\right)^2}}=\frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3 \end{aligned} $

We have, $Y=\sin a x+\cos b x$

$ \begin{aligned} & y^{\prime}=a \cos a x-b \sin b x \\\\ & y^{\prime \prime}=-a^2 \sin a x-b^2 \cos b x \\\\ & \therefore y^{\prime \prime}+b^2 y=-a^2 \sin a x-b^2 \cos b x \\\\ & +b^2(\sin a x+\cos b x)=\left(b^2-a^2\right) \sin a x \end{aligned} $

We have,

$ \begin{aligned} & S=t^3-t^2-t+3 \\\\ & V=\frac{d s}{d t}=3 t^2-2 t-1 \end{aligned} $

When it comes rest $\Rightarrow V=0$

$ \begin{aligned} \Rightarrow \quad & 3 t^2-2 t-1=0 \\\\ & 3 t^2-3 t+t-1=0 \\\\ & 3 t(t-1)+1(t-1)=0 \\\\ & t=1 \end{aligned} $

So, distance travelled by particle when it comes to rest is

$ S=(l)^3-(1)^2-1+3=2 \mathrm{~m} $

We have, $f(x)=\sqrt{x}$

$ (x \geq 0) $

$ \begin{aligned}and\,\, g(x) & =1+x^2 \\ f \circ g(x) & =\sqrt{x^2+1} \\ f^{\prime} g^{\prime}(x) & =\frac{1}{2 \sqrt{x^2+1}} \times 2 x \\ f \circ g^{\prime}(1) & =\frac{1}{2 \sqrt{1^2+1}} \times 2 \times 1=\frac{1}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} & \text { (i) } x=a(\theta-\sin \theta), y=a(1-\cos \theta) \\ & \begin{aligned} & \frac{d x}{d \theta}=a(1-\cos \theta), \frac{d y}{d \theta}=a \sin \theta \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\sin \theta}{1-\cos \theta} \\ &=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}}=\cot \frac{\theta}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}}=\cot \frac{\pi}{6}=\sqrt{3}\\ &\text { }\\ &\begin{aligned}(ii) & x=3 \cos \theta-2 \cos ^3 \theta \\ & \frac{d x}{d \theta}=-3 \sin \theta+6 \cos ^2 \theta \sin \theta \\ & y=3 \sin \theta-2 \sin ^3 \theta \\ & \frac{d y}{d \theta}=3 \cos \theta-6 \sin ^2 \theta \cos \theta \\ & \frac{d y}{d x}=\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(\cos ^2 \theta-1\right)}=\cot \theta \\ \left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}} & =\cot \frac{\pi}{3}=\frac{1}{\sqrt{3}} \\ \text { (iii) } x & =3 \cos \theta-\cos ^3 \theta \\ \frac{d x}{d \theta} & =-3 \sin \theta+3 \cos ^2 \theta \sin \theta \\ y & =3 \sin \theta-\sin ^3 \theta \\ \frac{d y}{d \theta} & =3 \cos \theta-3 \sin ^2 \theta \cos \theta \\ \frac{d y}{d x} & =\frac{3 \cos \theta\left(1-\sin ^2 \theta\right)}{3 \sin \theta\left(\cos ^2 \theta-1\right)} \\ & =\cot \theta \times \frac{\cos ^2 \theta}{-\left(\sin ^2 \theta\right)}=\cot ^3 \theta \end{aligned} $

$ \begin{aligned} \left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}} & =-\cot ^3 \frac{\pi}{3} \\ & =-\left(\frac{1}{\sqrt{3}}\right)^3=-\frac{1}{3 \sqrt{3}} \end{aligned} $

(iv) $\frac{d y}{d x}=\frac{a \sec ^2 \theta}{\frac{a}{\sin \theta} \cos \theta}=\frac{\sin \theta}{\cos ^3 \theta}$

At $\theta=\frac{\pi}{3}$,

$ \frac{d y}{d x}=\frac{\sqrt{3} / 2}{1 / 8}=4 \sqrt{3} $

Hence, option (c) is correct.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=(\sin x) x \\ & \frac{d y}{d x}=x \cos x+\sin x \\ & \text { Now, } \frac{\frac{d y}{d x}-\frac{y}{x}}{x \frac{d y}{d x}-y}=\frac{\frac{d y}{d x}-\frac{y}{x}}{x\left(\frac{d y}{d x}-\frac{y}{x}\right)}=\frac{1}{x}=\frac{1}{\alpha}=1 \\ & \therefore \quad \alpha=1 \end{aligned} \end{aligned} $

Given,

$ 2 x^2-3 x y+y^2+x+2 y-8=0 $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & 4 x-3\left(x \frac{d y}{d x}+y\right)+2 y \frac{d y}{d x}+1+2 \frac{d y}{d x}-0=0 \\ & \Rightarrow \frac{d y}{d x}(-3 x+2 y+2)+(4 x-3 y+1)=0 \\ & \Rightarrow(-3 x+2 y+2) d y-(-4 x+3 y-1) d x=0 \\ & \therefore f(x, y)=-3 x+2 y+2 \text { and } \\ & \quad g(x, y)=-4 x+3 y-1 \end{aligned} $

$ \begin{aligned}Now, \frac{g(2,2)}{f(1,1)} & =\frac{-4(2)+3(2)-1}{-3(1)+2(1)+2} \\ & =\frac{-8+6-1}{-3+2+2} \\ & =\frac{-3}{1}=-3 \end{aligned} $

Given, $f(x)=e^x$ and $h(x)=f \circ f(x)$

$ \begin{aligned} h(x) & =f \circ f(x)=f(f(x))=f\left(e^x\right)=e^{e^x} \\ \Rightarrow \log h(x) & =\log e^{e^x}=e^x \end{aligned} $

On differentiating w.r.t. $x$, we get

$ \frac{1}{h(x)} \cdot h^{\prime}(x)=e^x \Rightarrow \frac{h^{\prime}(x)}{h(x)}=\log h(x) $

$\sin y=\sin 3 t \Rightarrow \cos y \frac{d y}{d t}=3 \cos 3 t$

and $x=\sin t \Rightarrow d x / d t=\cos t$

$ \begin{aligned}Then, \frac{d y}{d x} & =\frac{3 \cos 3 t}{\cos y \cdot \cos t} \\ & =\frac{3 \cos 3 t}{\cos 3 t \cos t} \quad[\because \sin y=\sin 3 t] \\ & =\frac{3}{\cos t} \quad\left[\because \cos \theta=\sqrt{1-\sin ^2 \theta}\right] \\ & =\frac{3}{\sqrt{1-\sin ^2 t}} \\ & =\frac{3}{\sqrt{1-x^2}} \quad \quad[\because x=\sin t] \end{aligned} $

$ \begin{aligned} &\text {Given, }\\ &\begin{aligned} & f(x)=\sqrt{\log \left(x^2+x+1\right)+\sqrt{\cosh (2 x-3)}} \\ & f(x)=\left[\log \left(x^2+x+1\right)+(\cosh (2 x-3))^{1 / 2}\right]^{1 / 2} \\ & f^{\prime}(x)=\frac{1}{2}\left[\log \left(x^2+x+1\right)\right. \\ & \left.\quad+(\cosh (2 x-3))^{1 / 2}\right]^{-1 / 2} \\ & {\left[\frac{d}{d x} \log \left(x^2+x+1\right)+\frac{d}{d x}(\cosh (2 x-3))^{1 / 2}\right]} \\ & =\frac{1}{2}\left(\log \left(x^2+x+1\right)+(\cosh (2 x-3))^{1 / 2}\right)^{-1 / 2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \quad\left[\begin{array}{r} \frac{2 x+1}{x^2+x+1}+\frac{1}{2}(\cosh (2 x-3))^{-1 / 2} \\ \times \sinh (2 x-3) \times 2 \end{array}\right] \\ & f^{\prime}(0)=\frac{1}{2}\left[\log (0+0+1)+(\cosh (-3))^{1 / 2}\right]^{-1 / 2} \\ & \quad\left[\frac{1}{1}+\frac{1}{2}(\cosh (-3))^{-1 / 2} \times \sinh (-3) \times 2\right] \\ & =\frac{1}{2}\left[0+(\cosh 3)^{1 / 2}\right]^{-1 / 2} \times\left[1+(\cosh 3)^{-1 / 2}\right.\times(-\sin h 3)] \\ & =\frac{1}{2}(\cosh 3)^{-1 / 4}\left(1-\frac{\sinh 3}{(\cosh 3)^{1 / 2}}\right) \\ & =\frac{\cosh (3)-\sinh (3)}{2(\cosh (3))^{3 / 4}} \end{aligned} $

Given, $x=\cos ^3 \theta-\sin ^3 \theta$ and $y=(\cos \theta)^{1 / 3}-(\sin \theta)^{1 / 3}$

On differentiating $x$ and $y$ separately w.r.t. θ, we get

$ \begin{aligned} & \frac{d x}{d \theta}=3 \cos ^2 \theta(-\sin \theta)-3 \sin ^2 \theta(\cos \theta) \\ & \Rightarrow \frac{d x}{d \theta}=-3 \cos \theta \sin \theta(\cos \theta+\sin \theta)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \frac{d y}{d \theta}=\frac{1}{3}(\cos \theta)^{-2 / 3}(-\sin \theta) \\ & \left.\qquad-\frac{1}{3}(\sin \theta)^{-2 / 3}\right] \cos \theta \end{aligned} $

$ \begin{aligned} & =-\frac{1}{3}\left[\frac{\sin \theta}{(\cos \theta)^{2 / 3}}+\frac{\cos \theta}{(\sin \theta)^{2 / 3}}\right] \\ & =-\frac{1}{3} \sin \theta \cos \theta\left[(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Now, $\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$

$ \begin{aligned} = & \frac{-\frac{1}{3} \sin \theta \cos \theta\left[(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}\right]}{-3 \sin \theta \cos \theta[\sin \theta+\cos \theta]} \\ \frac{d y}{d x}= & \frac{1}{9}\left[\frac{(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}}{(\sin \theta+\cos \theta)}\right] \end{aligned} $

At $\quad \theta=\frac{\pi}{4}$

$ \begin{aligned} \left.\frac{d y}{d x}\right|_{\theta=\pi / 4} & =\frac{1}{9}\left[\frac{\left(\cos \frac{\pi}{4}\right)^{-5 / 3}+\left(\sin \frac{\pi}{4}\right)^{-5 / 3}}{\sin \frac{\pi}{4}+\cos \frac{\pi}{4}}\right] \\ & =\frac{1}{9}\left[\frac{\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}+\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}\right] \\ & =\frac{1}{9}\left[\frac{2\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}}{2\left(\frac{1}{\sqrt{2}}\right)}\right]=\frac{1}{9}\left(\frac{1}{\sqrt{2}}\right)^{-\frac{5}{3}-1} \\ & =\frac{1}{9}\left(\frac{1}{\sqrt{2}}\right)^{-8 / 3} \\ & =\frac{(\sqrt{2})^{8 / 3}}{9}=\frac{(2)^{8 / 3 \times 1 / 2}}{9}=\frac{2^{4 / 3}}{9} \\ & =\frac{2 \sqrt[3]{2}}{9} \end{aligned} $

Given,

$ 2 x^2+3 x y-y^2+4 x-5 y+6=0 $

On differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} & 4 x+3 y+3 x \frac{d y}{d x}-2 y \frac{d y}{d x}+4-\frac{5 d y}{d x}=0 \\ & \Rightarrow \quad-3 x \frac{d y}{d x}+2 y \frac{d y}{d x}+5 \frac{d y}{d x}=4 x+3 y+4 \\ & \Rightarrow \quad(-3 x+2 y+5) \frac{d y}{d x}=4 x+3 y+4 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{4 x+3 y+4}{-3 x+2 y+5} \end{aligned} $

$ \begin{aligned} \left(\frac{d y}{d x}\right)_{(1,-2)} & =\frac{4 \times 1+3 \times-2+4}{-3 \times 1+2 \times-2+5} \\ & =\frac{4-6+4}{-3-4+5}=\frac{2}{-2}=-1 \end{aligned} $

$ \begin{aligned} & f(x)=|x-1|+|x-2| \\ & f(x)=\left\{\begin{array}{cc} -2 x+3, & x<1 \\ 1, & 1 \leq x<2 \\ 2 x-3, & 2 \leq x \end{array}\right. \end{aligned} $

$ \begin{aligned} & \Rightarrow f^{\prime}(x)=\left\{\begin{array}{cc} -2, & x<1 \\ 0, & 1 \leq x<2 \\ 2, & 2 \leq x \end{array}\right. \\ & \Rightarrow f^{\prime}(-2023)=-2 \Rightarrow f^{\prime}\left(\frac{2024}{2023}\right)=0 \\ & \Rightarrow f^{\prime}(2023)=2 \\ & \therefore f^{\prime}(-2023)+f^{\prime}\left(\frac{2024}{2023}\right)+f^{\prime}(2023) \\ & =-2+0+2=0 \end{aligned} $

$ \begin{aligned} & \text { } f(x)=\frac{e^{2 x}-e^{-2 x}}{e^{3 x}+e^{-3 x}} \\ & f^{\prime}(x)=\frac{\left(e^{3 x}+e^{-3 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{-\left(e^{2 x}-e^{-2 x}\right)\left(3 e^{3 x}-3 e^{-3 x}\right)} \\ & \left(e^{3 x}+e^{-3 x}\right)^2 \\ & f^{\prime}(0)=\frac{2(4)-(0)}{2^2}=2 \end{aligned} $

$f(x)=x^{\tan x}+(\tan x)^x$

Let $\quad u=x^{\tan x}$ and $v=(\tan x)^x$

$ \begin{aligned} \log u & =\tan x \log x \\ \frac{1}{u} \frac{d u}{d x} & =\frac{\tan x}{x}+\sec ^2 x \log x \\ \frac{d u}{d x} & =x^{\tan x}\left(\frac{\tan x}{x}+\sec ^2 x \log x\right) \end{aligned} $

and $\log v=x \log \tan x$

$ \begin{aligned} \frac{1}{v} \frac{d v}{d x}= & \frac{x \sec ^2 x}{\tan x}+\log \tan x \\ \frac{d v}{d x}= & (\tan x)^x\left(\frac{x \sec ^2 x}{\tan x}+\log \tan x\right) \\ f^{\prime}(x)= & \frac{d u}{d x}+\frac{d v}{d x} \\ = & x^{\tan x}\left(\frac{\tan x}{x}+\sec ^2 x \cdot \log x\right) \\ & +(\tan x)^x\left(\frac{x \sec ^2 x}{\tan x}+\log \tan x\right) \end{aligned} $

$ \begin{aligned} f^{\prime}\left(\frac{\pi}{4}\right) & =\frac{\pi}{4}\left(\frac{4}{\pi}+2 \log \frac{\pi}{4}\right)+\left(2 \cdot \frac{\pi}{4}+\log 1\right) \\ & =1+\frac{\pi}{2} \log \frac{\pi}{4}+\frac{\pi}{2} \\ & =1+\frac{\pi}{2}\left(1+\log \frac{\pi}{4}\right) \\ & =1+\frac{\pi}{2}\left(\log e+\log \frac{\pi}{4}\right) \\ & =1+\frac{\pi}{2} \log \frac{e \pi}{4} \end{aligned} $

Given, $\sec \left(\log _2 y^2\right)=\operatorname{cosec}\left(\log _2 x^2\right)$

$ \begin{aligned} & \Rightarrow \sec \left(\log _2 y^2\right)=\sec \left(90^{\circ}-\log _2 x^2\right) \\ & \Rightarrow \quad \log _2 y^2=90^{\circ}-\log _2 x^2 \\ & \Rightarrow \quad 2 \log _2 y=90^{\circ}-2 \log _2 x \end{aligned} $

Differentiating on both sides w.r.t. $x$, we get

$ 2 \cdot \frac{1}{y} \frac{1}{\ln 2} \frac{d y}{d x}=-\frac{2}{x} \frac{1}{\ln 2} \Rightarrow \frac{d y}{d x}=-\frac{y}{x} $

Given, $e^x=y+\sqrt{y^2-1}$

Differentiating on both side w.r.t. $x$, we get

$ \begin{aligned} & \quad e^x=\frac{d y}{d x}+\frac{2 y}{2 \sqrt{y^2-1}} \frac{d y}{d x} \\ & \Rightarrow \quad e^x=\frac{d y}{d x}\left(\frac{\sqrt{y^2-1}+y}{\sqrt{y^2-1}}\right) \\ & \Rightarrow \quad e^x=\frac{d y}{d x}\left(\frac{e^x}{\sqrt{y^2-1}}\right) \Rightarrow \frac{d y}{d x}=\sqrt{y^2-1} \\ & \text { Now, } e^{-x}=\frac{1}{y+\sqrt{y^2-1}}=y-\sqrt{y^2-1} \end{aligned} $

Now, $\frac{e^x-e^{-x}}{2}=\sqrt{y^2-1}$

So, $\frac{d y}{d x}=\frac{e^x-e^{-x}}{2}=\sin \mathrm{h} x$

Given:

$ x = \log p $

Differentiating $ x $ with respect to $ p $ gives:

$ \frac{dx}{dp} = \frac{1}{p} $

We also have:

$ y = \frac{1}{p} $

Differentiating $ y $ with respect to $ p $ yields:

$ \frac{dy}{dp} = -\frac{1}{p^2} $

Next, we need to find $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = \frac{\frac{dy}{dp}}{\frac{dx}{dp}} = \frac{-\frac{1}{p^2}}{\frac{1}{p}} = -\frac{1}{p} $

Since $ p = e^x $, we can substitute:

$ \frac{dy}{dx} = -\frac{1}{e^x} = -e^{-x} $

Given the equation:

$ \tan y = \cot \left(\frac{\pi}{4} - x\right) $

We need to differentiate both sides with respect to $ x $.

Starting with the left side:

$ \frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx} $

For the right side, we differentiate:

$ \frac{d}{dx}\left(\cot \left(\frac{\pi}{4} - x\right)\right) = -\csc^2\left(\frac{\pi}{4} - x\right) \cdot \left(-1\right) = \csc^2\left(\frac{\pi}{4} - x\right) $

Setting the derivatives equal gives:

$ \sec^2 y \cdot \frac{dy}{dx} = \csc^2\left(\frac{\pi}{4} - x\right) $

Thus, solving for $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{\csc^2\left(\frac{\pi}{4} - x\right)}{\sec^2 y} $

Next recognize that:

$ \sec^2 y = 1 + \tan^2 y $

Given that $\tan y = \cot \left(\frac{\pi}{4} - x\right)$, we recognize that:

$ \tan\left(\frac{\pi}{4} - x\right) = \frac{1}{\tan y} $

Therefore:

$ \frac{dy}{dx} = \frac{\csc^2\left(\frac{\pi}{4} - x\right)}{1 + \tan^2\left(\frac{\pi}{4} + x\right)} $

Given the expressions:

$ x = 3 \sqrt{2} \cos^3 \theta $

and

$ y = 4 \tan^2 \theta, $

we want to find the derivative $\left(\frac{d y}{d x}\right)_{\theta=\pi / 4}$.

To do this, we will use the chain rule to compute $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}$, and then find $\frac{d y}{d x}$ by dividing the two derivatives.

First, compute $\frac{d y}{d \theta}$:

$ y = 4 \tan^2 \theta $

Differentiating with respect to $\theta$:

$ \frac{d y}{d \theta} = 4 \cdot 2 \tan \theta \sec^2 \theta = 8 \tan \theta \sec^2 \theta. $

Next, compute $\frac{d x}{d \theta}$:

$ x = 3 \sqrt{2} \cos^3 \theta $

Differentiating with respect to $\theta$ and using the chain rule:

$ \frac{d x}{d \theta} = 3 \sqrt{2} \cdot 3 \cos^2 \theta (-\sin \theta) = -9 \sqrt{2} \cos^2 \theta \sin \theta. $

Now, use these to find $\frac{d y}{d x}$:

$ \frac{d y}{d x} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} = \frac{8 \tan \theta \sec^2 \theta}{-9 \sqrt{2} \cos^2 \theta \sin \theta}. $

Evaluate this derivative at $\theta = \frac{\pi}{4}$:

At $\theta = \frac{\pi}{4}$:

$\tan \frac{\pi}{4} = 1$

$\sec \frac{\pi}{4} = \sqrt{2}$

$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute these values in:

$ \frac{d y}{d x}\bigg|_{\theta=\frac{\pi}{4}} = \frac{8 \cdot 1 \cdot (\sqrt{2})^2}{-9 \sqrt{2} \left(\frac{1}{\sqrt{2}}\right)^2 \cdot \frac{1}{\sqrt{2}}} = \frac{8 \cdot 2}{-9 \sqrt{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}}. $

Simplifying gives:

$ = \frac{16}{-9 \sqrt{2} \cdot \frac{1}{2\sqrt{2}}} = \frac{-32}{9}. $

Thus, $\left(\frac{d y}{d x}\right)_{\theta=\pi / 4} = -\frac{32}{9}$.

We need to find the derivative of

$ y(x) = \frac{1-x^2}{1+x^2} $

with respect to

$ u(x) = \frac{2x}{1+x^2} $

at $ x = 2 $. In other words, we want to compute

$ \frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}} $

evaluated at $ x = 2 $.

Here’s how to do it step by step:

Differentiate $ y(x) $ with respect to $ x $:

We have

$ y(x) = \frac{1-x^2}{1+x^2}. $

Using the quotient rule, where if $ f(x) = \frac{v(x)}{w(x)} $ then

$ f'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2}, $

let:

$ v(x) = 1-x^2 $ so that $ v'(x) = -2x $,

$ w(x) = 1+x^2 $ so that $ w'(x) = 2x $.

Therefore,

$ y'(x) = \frac{(-2x)(1+x^2) - (1-x^2)(2x)}{(1+x^2)^2}. $

Simplify the numerator:

$ \begin{aligned} (-2x)(1+x^2) - (1-x^2)(2x) &= -2x(1+x^2) - 2x(1-x^2) \\ &= -2x\Big[(1+x^2) + (1-x^2)\Big] \\ &= -2x(2) \\ &= -4x. \end{aligned} $

Thus,

$ y'(x) = \frac{-4x}{(1+x^2)^2}. $

Differentiate $ u(x) $ with respect to $ x $:

We have

$ u(x) = \frac{2x}{1+x^2}. $

Again, using the quotient rule:

$ v(x) = 2x $ so that $ v'(x) = 2 $,

$ w(x) = 1+x^2 $ so that $ w'(x) = 2x $.

Therefore,

$ u'(x) = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2}. $

Simplify the numerator:

$ \begin{aligned} 2(1+x^2) - 4x^2 &= 2 + 2x^2 - 4x^2 \\ &= 2 - 2x^2 \\ &= 2(1-x^2). \end{aligned} $

Thus,

$ u'(x) = \frac{2(1-x^2)}{(1+x^2)^2}. $

Compute $ \frac{dy}{du} $:

By the chain rule:

$ \frac{dy}{du} = \frac{y'(x)}{u'(x)} = \frac{\frac{-4x}{(1+x^2)^2}}{\frac{2(1-x^2)}{(1+x^2)^2}}. $

Notice that the factor $ (1+x^2)^2 $ cancels out, so:

$ \frac{dy}{du} = \frac{-4x}{2(1-x^2)} = \frac{-2x}{1-x^2}. $

Evaluate at $ x = 2 $:

Substitute $ x = 2 $:

$ \frac{dy}{du}\Bigg|_{x=2} = \frac{-2(2)}{1-2^2} = \frac{-4}{1-4} = \frac{-4}{-3} = \frac{4}{3}. $

Thus, the derivative of $ \frac{1-x^2}{1+x^2} $ with respect to $ \frac{2x}{1+x^2} $ at $ x=2 $ is $ \frac{4}{3} $.

This matches Option B.

To find the value of $2a - 3b$, consider the function $y = e^{a + bx^2}$.

Given Point on the Curve:

The point $P(1, 1)$ lies on the curve. At $x = 1$, $y = 1$:

$ 1 = e^{a + b} \implies a + b = 0 $

Slope of the Tangent:

The slope of the tangent at $(1, 1)$ is given as $-2$. We find the derivative:

$ \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) $

Evaluate the Derivative at $x = 1$:

At $x = 1$:

$ \left. \frac{dy}{dx} \right|_{(1,1)} = e^{a + b} \cdot 2b \cdot 1 = e^{a + b} \cdot 2b = -2 $

Since $e^{a+b} = 1$, we have:

$ 1 \cdot 2b = -2 \implies 2b = -2 \implies b = -1 $

Find $a$:

Substituting $b = -1$ into $a + b = 0$:

$ a - 1 = 0 \implies a = 1 $

Calculate $2a - 3b$:

$ 2a - 3b = 2(1) - 3(-1) = 2 + 3 = 5 $

Thus, the value of $2a - 3b$ is 5.

$ f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\left\{\frac{4}{5} \sin (p x)-\frac{3}{5} \cos (p x)\right\} $

Let $\cos \alpha=\frac{4}{5}$, then $\sin \alpha=\frac{3}{5}$

Here, $\frac{\sin \alpha}{\cos \alpha}=\frac{3}{4} \Rightarrow \alpha=\tan ^{-1}\left(\frac{3}{4}\right)$

$ \begin{aligned} f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\{\cos \alpha \sin (p x) & \\ & -\sin \alpha \cos (p x)\} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\sum_{p=1}^7 p^2 \sin ^{-1}\{\sin (p x-\alpha)\} \\ & =\sum_{p=1}^7 p^2(p x-\alpha) \\ & =\sum_{p=1}^7 p^3 x-p^2 \alpha \\ & =x \cdot \sum_{p=1}^7 p^3-\alpha \sum_{p=1}^7 p^2 \\ & =x\left[\frac{7(7+1)}{2}\right]^2-\alpha\left[\frac{7(7+1)(2 \times 7+1)}{6}\right] \\ & =x(28)^2-\alpha[7 \cdot 4 \cdot 5] \\ & =784 x-140 \alpha \\ f(x) & =784 x-140 \tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}\\ &\text { On differentiating w.r.t. } x \text {, we get }\\ &f^{\prime}(x)=784 \times 1-0=784 \end{aligned} $

Given, $y=\frac{a x+b}{c x+d}$

$ \begin{aligned} \Rightarrow & & y c x+y d & =a x+b \\ \Rightarrow & & y x c-a x & =b-y d \\ \Rightarrow & & x & =\frac{b-y d}{c y-a} \end{aligned} $

On differentiating w.r.t. $y$,

$ \begin{aligned} \frac{d x}{d y} & =\frac{(c y-a)(-d)-(b-y d)(c)}{(c y-a)^2} \\ & =\frac{-c d y+a d-b c+c d y}{(a-c y)^2} \end{aligned} $

Hence, $\frac{d x}{d y}=\frac{a d-b c}{(a-c y)^2}$

$ \begin{aligned} & \text { } \\ & \begin{aligned} \text { } & x^2+y^2=t-\frac{1}{t} \\ & x^4+y^4=t^2+\frac{1}{t^2} \\ & \left(x^2+y^2\right)^2=\left(t-\frac{1}{t}\right)^2 \\ \Rightarrow & x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}-2 \\ \Rightarrow & t^2+\frac{1}{t^2}+2 x^2 y^2=t^2+\frac{1}{t^2}-2 \\ \Rightarrow & 2 x^2 y^2=-2 \\ \Rightarrow & x^2 y^2=-1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { On differentiating w.r.t. } x \text {, we get }\\ &\begin{gathered} 2 x y^2+x^2\left(2 y y^{\prime}\right)=0 \\ \Rightarrow y^{\prime}=\frac{-2 x y^2}{2 x^2 y} \Rightarrow \frac{d y}{d x}=\frac{-y}{x} \end{gathered} \end{aligned} $