Differentiation

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} $

To find the derivative $\frac{dy}{dx}$ for $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$, we use the formula for the derivative of the inverse hyperbolic sine function:

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

Applying this to our function:

Identify $u = \frac{1-x}{1+x}$.

Differentiate $u$ with respect to $x$:

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

Now apply the chain rule:

$ \frac{dy}{dx} = \frac{1}{\sqrt{1+\left(\frac{1-x}{1+x}\right)^2}} \times \frac{du}{dx} $

Simplify the expression:

Compute $1 + \left(\frac{1-x}{1+x}\right)^2$:

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

Note that $(1+x)^2 + (1-x)^2 = 2(1+x^2)$

Plug this into the derivative formula:

$ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{2(1+x^2)}{(1+x)^2}}} \times \frac{-2}{(1+x)^2} $

Simplify further to get the final derivative:

$ = \frac{-2}{\sqrt{2(1+x^2)} \cdot (1+x)} = \frac{-\sqrt{2}}{|1+x| \sqrt{1+x^2}} $

Thus, the derivative $\frac{dy}{dx}$ is:

$ \frac{-\sqrt{2}}{|1+x| \sqrt{1+x^2}} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & y=(x-1)(x+2)\left(x^2+5\right)\left(x^4+8\right) \\ & \Rightarrow y=\left(x^4+x^3+3 x^2+5 x-10\right)\left(x^4+8\right) \\ & \Rightarrow \frac{d y}{d x}=\left(x^4+x^3+3 x^2+5 x-10\right)\left(4 x^3\right) \\ & \quad+\left(4 x^3+3 x^2+6 x+5\right)\left(x^4+8\right) \\ & \begin{aligned} \therefore \lim _{x \rightarrow-1}\left(\frac{d y}{d x}\right) & =(1-1+3-5-10)(-4) \\ & \quad+(-4+3-6+5)(1+8) \\ = & (-12)(-4)+(-2)(9) \\ = & 48-18=30 \end{aligned} \end{aligned} \end{aligned} $

To find the expression for $(1+4x^2)^2 y'' - 16$, we start with

$ y = \left(\tan^{-1} 2x\right)^2 + \left(\cot^{-1} 2x\right)^2. $

Using the identity $\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u$, we can rewrite the equation as

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

Simplifying,

$ y = 2\left(\tan^{-1} 2x\right)^2 + \frac{\pi^2}{4} - \pi \tan^{-1} 2x. $

Next, we differentiate $y$ with respect to $x$:

$ y' = \frac{4 \tan^{-1} 2x}{1 + 4x^2} \cdot 2 - \frac{\pi}{1 + 4x^2} \cdot 2. $

This simplifies to

$ y' = \frac{8 \tan^{-1} 2x - 2\pi}{1 + 4x^2}. $

Multiplying both sides by $(1 + 4x^2)$, we get

$ (1 + 4x^2) y' = 8 \tan^{-1} 2x - 2\pi. $

Differentiating this equation with respect to $x$ gives:

$ (1 + 4x^2) y'' + y'(8x) = \frac{16}{1 + 4x^2}, $

Rearranging this result, we find:

$ (1 + 4x^2)^2 y'' + 8x y'(1 + 4x^2) = 16. $

Thus,

$ (1 + 4x^2)^2 y'' - 16 = -8x y'(1 + 4x^2). $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \begin{array}{l} y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2} \\ \\ +\tan ^{-1} \frac{x}{1+12 x^2} \\ \Rightarrow \frac{d y}{d x}=\frac{1}{1+\left(\frac{x}{1+2 x^2}\right)^2} \cdot \frac{\left(1+2 x^2\right) \cdot 1-x \cdot 4 x}{\left(1+2 x^2\right)^2} \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & +\frac{1}{1+\left(\frac{x}{1+6 x^2}\right)^2} \times \frac{\left(1+6 x^2\right) \cdot 1-\times(12,x)}{\left(1+6 x^2\right)^2} \\ & +\frac{1}{1+\left(\frac{x}{1+12 x^2}\right)^2} \cdot \frac{\left(1+12 x^2\right) \cdot 1-x \cdot 24x}{\left(1+12 x^2\right)^2} \end{aligned}\\ &\text { put, } x=\frac{1}{2} \text { and simplify }\\ &\left.\frac{d y}{d x}\right|_{x=\frac{1}{2}}=0 \end{aligned} $

Given that

$ f(x)=5 \cos ^3 x-3 \sin ^2 x $

and $g(x)=4 \sin ^3 x+\cos ^2 x$

Now, $\frac{d f(x)}{d x}=-15 \cos ^2 x \cdot \sin x-6 \sin x \cos x $

$ \begin{aligned} & =-\frac{15 \cos ^2 x \sin x-6 \sin x \cos x}{12 \sin ^2 x \cos x-2 \cos x \sin x} \\ & =\frac{-\cos x \sin x(15 \cos x+6)}{-\cos x \sin x(-12 \sin x+2)} \\ & =-\left(\frac{15 \cos x+6}{12 \sin x-2}\right) \end{aligned} $

Given the series:

$ y = 1 + x + x^2 + x^3 + \ldots $

and noting that $ |x| < 1 $, we recognize this as a geometric series. The sum of this series can be expressed in closed form as:

$ y = \frac{1}{1-x} $

Now, let's find the first derivative of $ y $ with respect to $ x $:

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

Next, differentiate $ y' $ to find the second derivative $ y'' $:

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

Since we have $ y = \frac{1}{1-x} $, it follows that:

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

This can be rewritten using the previously calculated derivatives:

$ y'' = 2y \cdot y' $

Thus, the second derivative of $ y $, $ y'' $, is equal to:

$ y'' = 2 y \cdot y' $

Given the expression:

$ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}} $

We will evaluate $\frac{d^2 y}{d x^2}$ at the point $(\pi, 1)$.

Step 1: Determine $ y $

Squaring both sides of the equation:

$ y^2 = \sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}} $

Therefore, from the self-similarity of the expression, we have:

$ y^2 - \sin x = y \quad \Rightarrow \quad y^2 - y = \sin x $

Step 2: First Derivative $\frac{d y}{d x}$

Differentiating both sides with respect to $x$:

$ 2y \frac{d y}{d x} - \cos x = \frac{d y}{d x} $

Rearranging, we get:

$ \frac{d y}{d x} (2y - 1) = \cos x \quad \Rightarrow \quad \frac{d y}{d x} = \frac{\cos x}{2y - 1} $

Step 3: Second Derivative $\frac{d^2 y}{d x^2}$

Differentiating the equation for $\frac{d y}{d x}$ again:

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

Rearranging:

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

Simplifying:

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

Thus:

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

Final Calculation at $(\pi, 1)$

At the point $(\pi, 1)$:

$\sin \pi = 0$

$\cos \pi = -1$

$y = 1$

Substituting these values into the formula for the second derivative:

$ \frac{d^2 y}{d x^2} = \frac{0 - \frac{2 \times (-1)^2}{(2 \times 1 - 1)^2}}{1} = \frac{-2}{1} = -2 $

Therefore, $\frac{d^2 y}{d x^2}$ at $(\pi, 1)$ is $-2$.

We are given the function $ y = f(f(f(f(f(x))))) $ and need to find its derivative at $ x = 0 $, where we know $ f(0) = 0 $ and $ f^{\prime}(0) = 3 $.

To find $ y^{\prime}(0) $, apply the chain rule:

$ y^{\prime} = f^{\prime}(f(f(f(f(x))))) \cdot f^{\prime}(f(f(f(x)))) \cdot f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) $

Substituting $ x = 0 $, we find:

$ y^{\prime}(0) = f^{\prime}(f(f(f(f(0))))) \cdot f^{\prime}(f(f(f(0)))) \cdot f^{\prime}(f(f(0))) \cdot f^{\prime}(f(0)) \cdot f^{\prime}(0) $

Since $ f(0) = 0 $, this simplifies to:

$ y^{\prime}(0) = f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) $

Given $ f^{\prime}(0) = 3 $, this becomes:

$ y^{\prime}(0) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5 = 243 $

Thus, the derivative of $ y $ at $ x = 0 $ is $ 243 $.

To find $ (a, b) $ such that

$ \frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right) = ax + b $

we start by simplifying the expression in the numerator.

Simplification

The numerator $ 1 + x^2 + x^4 $ can be factored:

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

Substituting Back

Using the factorized form of the numerator, rewrite the original expression:

$ \frac{d}{d x}\left(\frac{(x^2-x+1)(x^2+x+1)}{1+x+x^2}\right) = ax + b. $

We notice that $ (x^2+x+1) $ cancels out with the denominator:

$ \frac{d}{d x}(x^2 - x + 1) = ax + b. $

Differentiation

Calculating the derivative of the simplified expression $ x^2 - x + 1 $:

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

Identifying Constants

Set the expression for derivative equal to $ ax + b $:

$ 2x - 1 = ax + b. $

By comparing coefficients, we get:

$ a = 2, \quad b = -1. $

Thus, the values are:

$ (a, b) = (2, -1). $

To find the rate of change of $ x^{\sin x} $ with respect to $ (\sin x)^x $, we proceed as follows:

Let $ y = x^{\sin x} $.

Taking the natural logarithm on both sides, we have:

$ \log y = \sin x \log x $

Differentiating with respect to $ x $, we get:

$ \frac{1}{y} \frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x} $

Let $ z = (\sin x)^x $.

Taking the logarithm similarly, we have:

$ \log z = x \log \sin x $

Differentiating with respect to $ x $, we have:

$ \frac{1}{z} \frac{dz}{dx} = \log \sin x + x \frac{\cos x}{\sin x} $

Simplifying this gives:

$ \frac{1}{z} \frac{dz}{dx} = \log \sin x + x \cot x $

Next, we find the derivative of $ y $ with respect to $ z $ by dividing the derivatives:

$ \frac{dz}{dy} = \frac{yddx}{zddx} = \frac{\cos x \log x + \frac{\sin x}{x}}{\log \sin x + x \cot x} $

Thus, the rate of change is:

$ \frac{dy}{dz} = \frac{x^{\sin x}}{(\sin x)^x} \left( \frac{\log x \cos x + \frac{1}{x} \sin x}{\log \sin x + x \cot x} \right) $

$y=\frac{\alpha x+\beta}{\gamma x+\delta}$ differentiation w.r.t. $x$, we get

$ y_1=\frac{\alpha \delta-\beta \gamma}{(\gamma x+\delta)^2} $

Again, differentiation with respect to $x$, we get

$ y_2=(\alpha \delta-\beta \gamma)\left(\frac{-2 \gamma}{(\gamma x+\delta)^3}\right) $

Again, differentiation with respect to $x$, we get

$ \begin{aligned} & y_3=(\alpha \delta-\beta \gamma)\left(\frac{6 \gamma^2}{(\gamma x+\delta)^4}\right) \\ & y_3=\frac{(\alpha \delta-\beta \gamma)}{(\gamma x+\delta)^2} \cdot \frac{6 \gamma}{(\gamma x+\delta)^2} \end{aligned} $

On squaring both sides of Eq. (ii), we get

$ \begin{aligned} & y_2^2=(\alpha \delta-\beta \gamma)^2 \cdot \frac{4 \gamma^2}{(\gamma x+\delta)^6} \\ & y_2^2=\frac{(\alpha \delta-\beta \gamma)}{(\gamma x+\delta)^2} \cdot \frac{4 \gamma^2}{(\gamma x+\delta)^4} \\ & y_2^2=\frac{(\alpha \delta-\beta \gamma)}{(\gamma \alpha+\delta)^2} \cdot \frac{(\alpha \delta-\beta \gamma) \cdot 6 \gamma^2}{(\gamma x+\delta)^4} \cdot \frac{4}{6} \end{aligned} $

Here, from Eqs. (i) and (iii)

$ y_2^2=y_1 y_3 \cdot\left(\frac{2}{3}\right) $

$ \begin{aligned} &\begin{aligned} & y_2^2=\frac{2}{3} y_1 y_3 \\ & y_2^2=\frac{2}{3} y_1 y_3 \end{aligned}\\ &\text { Now, } 2 y_1 y_3=3 y_2^2 \end{aligned} $

The following explanations clarify the derivatives being calculated:

(A) $\frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right]=\operatorname{sech}(x)$

For the inverse secant function, the derivative is given by:

$ \frac{d}{d x}\left[\sec ^{-1}(\theta)\right] = \frac{1}{|\theta| \sqrt{\theta^2 - 1}} $

Substituting $\theta = \cosh x$:

$ \frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right] = \frac{1}{|\cosh x| \sqrt{(\cosh x)^2 - 1}} \cdot \sinh x $

Since $|\cosh x| = \cosh x$ for all $x \in \mathbb{R}$ and $(\cosh x)^2 - 1 = \sinh^2 x$, we find:

$ \frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right] = \frac{\sinh x}{\cosh x \sinh x} = \frac{1}{\cosh x} = \operatorname{sech}(x) $

This statement is true.

(B) $\frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right]=\operatorname{sech} x$

The derivative of the inverse cosine function is:

$ \frac{d}{d x}\left[\cos ^{-1}(\theta)\right] = -\frac{1}{\sqrt{1 - \theta^2}} $

Substituting $\theta = \operatorname{sech} x$:

$ \frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right] = -\frac{\operatorname{sech} x \tanh x}{\sqrt{1 - (\operatorname{sech} x)^2}} $

Since $1 - \operatorname{sech}^2 x = \tanh^2 x$, we have:

$ \frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right] = \operatorname{sech} x $

This statement is true.

(C) $\frac{d}{d x}\left[\tan ^{-1}(\sinh x)\right]=\operatorname{sech} x$

The derivative of the inverse tangent function is:

$ \frac{d}{d x} \tan ^{-1}(\theta) = \frac{1}{1 + \theta^2} $

Substituting $\theta = \sinh x$:

$ \frac{d}{d x}\left[\tan ^{-1}(\sinh x)\right] = \frac{1}{1 + (\sinh x)^2} $

Knowing that $\cosh^2 x - \sinh^2 x = 1$, it simplifies to:

$ \frac{\cosh x}{1 + \sinh^2 x} = \operatorname{sech} x $

This statement is true.

(D) $\frac{d}{d x}\left[\tan ^{-1}\left(\tan \frac{x}{2}\right)\right]=\sec x$

The expression simplifies as:

$ \tan^{-1}\left(\tan \frac{x}{2}\right) = \frac{x}{2} \quad \text{for } x \in (-\pi, \pi) $

Thus, the derivative is:

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

And $\frac{1}{2} \neq \sec x$, rendering this statement false.

Given the functions $y = t^2 + t^3$ and $x = t - t^4$, we need to find $\frac{d^2 y}{d x^2}$ at $t = 1$.

First, find the derivatives with respect to $t$:

$ \frac{d y}{d t} = 2t + 3t^2 $

$ \frac{d x}{d t} = 1 - 4t^3 $

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

$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2t + 3t^2}{1 - 4t^3} $

To find $\frac{d^2 y}{d x^2}$, differentiate $\frac{d y}{d x}$ with respect to $t$ and multiply by $\frac{d t}{d x}$:

$ \frac{d^2 y}{d x^2} = \frac{\left(1 - 4t^3\right)(2 + 6t) - (2t + 3t^2)(-12t^2)}{\left(1 - 4t^3\right)^2} \cdot \frac{dt}{dx} $

Substitute $\frac{dt}{dx} = \frac{1}{\frac{d x}{d t}}$:

$ = \frac{(2 + 6t)(1 - 4t^3) + 12t^2(2t + 3t^2)}{\left(1 - 4t^3\right)^2} \cdot \frac{1}{1 - 4t^3} $

Evaluate at $t = 1$:

$ \frac{d^2 y}{d x^2} = \frac{(2 + 6)(1 - 4) + 12(2 + 3)}{(1 - 4)^2} \cdot \frac{1}{1 - 4} $

Simplify the expression:

$ = \frac{(-24 + 60)}{-27} = \frac{36}{-27} = -\frac{4}{3} $

Therefore, $\frac{d^2 y}{d x^2}$ at $t = 1$ is $-\frac{4}{3}$.

To find the second derivative of $ y = \tan (\log x) $, follow these steps:

First Derivative:

Start by differentiating $ y = \tan (\log x) $ with respect to $ x $. Apply the chain rule:

$ \frac{dy}{dx} = \frac{d}{dx}[\tan(\log x)] = \sec^2(\log x) \cdot \frac{d}{dx}[\log x] = \frac{\sec^2(\log x)}{x} $

Second Derivative:

To find the second derivative, apply the quotient rule to $\frac{dy}{dx} = \frac{\sec^2(\log x)}{x}$. The quotient rule is given by:

$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $

Here, let $ u = \sec^2(\log x) $ and $ v = x $.

Then,

$ \frac{du}{dx} = 2 \sec^2(\log x) \tan(\log x) \cdot \frac{1}{x} = \frac{2 \sec^2(\log x) \tan(\log x)}{x} $

$ \frac{dv}{dx} = 1 $

Substitute into the quotient rule:

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

Simplify the expression:

$ \frac{d^2y}{dx^2} = \frac{2 \sec^2(\log x) \tan(\log x) - \sec^2(\log x)}{x^2} $

$ = \frac{\sec^2(\log x) [2 \tan(\log x) - 1]}{x^2} $

Thus, the second derivative is:

$ \frac{d^2y}{dx^2} = \frac{\sec^2(\log x)[2 \tan(\log x) - 1]}{x^2} $

For $ x < 0 $, we need to find the derivative $\frac{d}{dx}\left[|x|^x\right]$.

Start by letting $ y = (-x)^x $. Since $ x < 0 $, we have $ |x| = -x $.

We proceed by taking the natural logarithm of both sides:

$ \log y = x \log (-x) $

Now, differentiate both sides with respect to $ x $:

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

Use the product rule to differentiate the right side:

$ \frac{d}{dx}(x \log (-x)) = \log (-x) + x \cdot \frac{d}{dx}(\log (-x)) $

Since $\frac{d}{dx}(\log (-x)) = \frac{-1}{-x} = \frac{1}{x}$, we have:

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

Therefore:

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

Multiplying both sides by $ y = (-x)^x $:

$ \frac{dy}{dx} = (-x)^x (\log (-x) - 1) $

Simplify to:

$ \frac{dy}{dx} = (-x)^x [1 + \log (-x)] $

To find the rate of change of $ y^2 $ with respect to $ x^2 $, we begin with the given function:

$ y = x - x^2 $

We introduce new variables $ u = y^2 $ and $ v = x^2 $. Thus:

$ u = (x - x^2)^2 $

First, we find the derivatives $ \frac{du}{dx} $ and $ \frac{dv}{dx} $:

$ \begin{aligned} \frac{du}{dx} & = 2(x - x^2)(1 - 2x) \\ \frac{dv}{dx} & = 2x \end{aligned} $

Now, the rate of change of $ y^2 $ with respect to $ x^2 $ is:

$ \frac{dy^2}{dx^2} = \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2(x - x^2)(1 - 2x)}{2x} = (1 - x)(1 - 2x) $

We evaluate this expression at $ x = 2 $:

$ (1 - 2)(1 - 4) = (-1)(-3) = 3 $

Thus, the derivative of $ y^2 $ with respect to $ x^2 $ at $ x = 2 $ is 3.

If $ y = f(x) $ is a thrice differentiable function and a bijection, we want to evaluate the expression $ \frac{d^2 x}{dy^2}\left(\frac{dy}{dx}\right)^3 + \frac{d^2 y}{dx^2} $.

First, we know:

$ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} $

To find the second derivative of $ x $ with respect to $ y $, we differentiate:

$ \frac{d^2 x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dy}\left(\frac{dy}{dx}\right)^{-1} $

Applying the chain rule:

$ = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dy}{dx}} $

Rearranging terms, we have:

$ = -\left(\frac{dx}{dy}\right)^2 \cdot \frac{d^2 y}{dx^2} \cdot \left(\frac{dx}{dy}\right) $

Thus, simplifying further:

$ = -\left(\frac{dx}{dy}\right)^3 \cdot \frac{d^2 y}{dx^2} $

This simplifies our original expression:

$ \frac{d^2 x}{dy^2} \left(\frac{dy}{dx}\right)^3 = -\frac{d^2 y}{dx^2} $

Therefore:

$ \frac{d^2 x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2 y}{dx^2} = 0 $

$\begin{aligned} & y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right) \\ & \begin{array}{l}\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x}\right)^2} \\ \left\{\frac{(3-2 \sin x)(-3 \cos x)-(2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}\right\} \\ =\frac{-9 \cos x+6 \sin x \cos x+4 \cos x-6 \sin x \cos x}{(3-2 \sin x)^2+(2-3 \sin x)^2} \\ =\frac{-5 \cos x}{9+4 \sin ^2 x-12 \sin x+4+9 \sin ^2 x-12 \sin x} \\ =\frac{-5 \cos x}{13+13 \sin ^2 x-24 \sin x}\end{array}\end{aligned}$

To determine $\frac{dy}{dx}$, we begin with the expressions given for $x$ and $y$.

$ x = 3\left[\sin t - \log\left(\cot \frac{t}{2}\right)\right] $

Differentiating $x$ with respect to $t$:

$ \begin{align*} \frac{dx}{dt} &= 3\left[\cos t - \frac{1}{\cot \frac{t}{2}} \cdot (-\csc^2 \frac{t}{2}) \cdot \frac{1}{2}\right] \\ &= 3\left[\cos t + \frac{1}{\sin t}\right] \\ &= 3\left[\frac{\cos t \sin t + 1}{\sin t}\right] \end{align*} $

Next, differentiating $y$ with respect to $t$:

$ y = 6\left[\cos t + \log\left(\tan \frac{t}{2}\right)\right] $

$ \begin{align*} \frac{dy}{dt} &= 6\left[-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2}\right] \\ &= 6\left[-\sin t + \frac{1}{\sin t}\right] \\ &= 6\left[\frac{-\sin^2 t + 1}{\sin t}\right] \end{align*} $

Now, to find $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6\left[\frac{-\sin^2 t + 1}{\sin t}\right]}{3\left[\frac{\cos t \sin t + 1}{\sin t}\right]} $

The $\sin t$ terms cancel out:

$ = \frac{6(-\sin^2 t + 1)}{3(\cos t \sin t + 1)} $

Simplifying further:

$ = \frac{2(-\sin^2 t + 1)}{\cos t \sin t + 1} $

Rewriting the numerator, $1 - \sin^2 t = \cos^2 t$:

$ = \frac{2\cos^2 t}{\cos t \sin t + 1} $

Therefore, the expression for $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{2\cos^2 t}{1 + \sin t \cos t} $

To find the length of the tangent at the point $ P\left(\frac{\pi}{4}\right) $ on the curve $ x^{2/3} + y^{2/3} = 2^{2/3} $, we can proceed as follows:

The given equation of the curve can be parameterized as:

$ x = 2 \cos^3 \theta, \quad y = 2 \sin^3 \theta $

For the point $ P $, we substitute $ \theta = \frac{\pi}{4} $:

$ x = 2 \cos^3 \frac{\pi}{4}, \quad y = 2 \sin^3 \frac{\pi}{4} $

Calculating the coordinates:

$ x = 2 \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}, \quad y = 2 \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} $

So the point is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

Next, differentiate the curve equation $ x^{2/3} + y^{2/3} = 2^{2/3} $ with respect to $ x $:

$ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 $

Solving for $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} $

At the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$,

$ \frac{dy}{dx} = -1 $

The formula to find the length of the tangent at a point is:

$ L = y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} $

Given $\frac{dx}{dy} = -1$, the length of the tangent at the point is calculated as:

$ L = \frac{1}{\sqrt{2}} \sqrt{1 + (-1)^2} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1 $

Thus, the length of the tangent is $1$.

1. Given the equation: $3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

2. Replace $x$ with $\frac{1}{x}$ in the original equation:
   $3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

3. Now, we have two equations:

$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

1. By adding the two equations, we can find $f(x)$:

$5f(x) = \frac{3}{x} - 2x - 10$

1. Now, let's differentiate both sides with respect to $x$:

$5f'(x) = -\frac{3}{x^2} - 2$

1. Now, we can find the values for $f(3)$ and $f'\left(\frac{1}{4}\right)$:

$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$

$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$

1. Finally, calculate the expression we are interested in :

$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$

$f''(x) = g''(x) + 6x$

$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$

$f'(1) = g'(1) + 3 + C$

$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$

$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$

$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$

$x = 2$

$f(2) = g(2) + 14 + C'$

$12 = 4 + 14 + C'$

$ \Rightarrow C' = - 6$

$ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6$

$f( - 2) = g( - 2) - 8 - 6 - 6$

$g( - 2) - f( - 2) = 20$

$f'(x) - g'(x) = 3{x^2} + 3$

$x \in ( - 1,1)$

$3{x^2} + 3 \in (0,6)$

$ \Rightarrow f'(x) - g'(x) \in (0,6)$

$f(x) - g(x) = {x^3} + 3x - 6$

At $x = - 1$

$|f( - 1) - g( - 1)| = 10$

$\therefore$ Option (4) is false.