Differentiation

$ \begin{aligned} &\text { Let } f^{\prime}(1)=a\\ &\begin{aligned} & f^{\prime}(2)=b \\ & f^{\prime}(3)=c \\ & f(x)=x^3+a x^2+b x+c \\ & f^{\prime}(x)=3 x^2+2 a x+b \\ & f^{\prime}(x)=a=3+2 a+b \Rightarrow a+b=3 \ldots . . (1)\\ & f^{\prime \prime}(x)=6 x+2 a \\ & f^{\prime \prime}(2)=12+2 a=\frac{b}{2} \Rightarrow 4 a-b=-24 \\ & f^{\prime \prime}(x)=6 \\ & f^{\prime \prime}(3)=c=6 \\ & \Rightarrow a=\frac{-27}{5}, b=\frac{12}{5} \\ & f^{\prime}(5)=75+10 a+b \\ & =75-54+\frac{12}{5}=21+\frac{12}{5}=\frac{117}{5} \end{aligned} \end{aligned} $

To solve for $ y(x) $, we start with the determinant of the given matrix:

$ \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} $

Using the formula for a $3 \times 3$ determinant, we expand along the first row:

$ f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} $

Calculating the smaller $2 \times 2$ determinants:

$ \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1 $

$ \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0 $

$ \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1 $

Substituting these into the determinant calculation gives:

$ f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1) $

$ f(x) = \sin x - (\sin x + \cos x + 1) $

$ f(x) = \sin x - \sin x - \cos x - 1 $

$ f(x) = -\cos x - 1 $

Now, calculate the derivatives:

First derivative $ f'(x) $:

$ f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x $

Second derivative $ \frac{d^2 f}{dx^2} $:

$ \frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x $

Finally, compute $ \frac{d^2 y}{dx^2} + y $:

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

Thus, $ \frac{d^2 y}{dx^2} + y $ is equal to $-1$.

$\begin{aligned} & \sin (x-y) f(2 x+2 y)=f(2 x-2 y) \sin (x+y) \\ & \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}=k(\text { say }) \\ & f(2 x+2 y)=k \sin (x+y) \\ & f(2 x)=5 \sin x \quad(\because y=0) \\ & f(x)=k \sin \frac{x}{2} \\ & f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2} \\ & f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1 \\ & f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2} \\ & f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \\ & 24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3 \end{aligned}$

$\begin{aligned} & x^2 f^{\prime}(x)-2 x f(x)=3 \\ & \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \\ & \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4} \end{aligned}$

Integrating both sides

$\begin{aligned} & \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \\ & \text { put } \mathrm{x}=1 \\ & 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2 \end{aligned}$

Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]$

$=39$

$\begin{aligned} &\log _e y=3 \sin ^{-1} x\\ &\begin{aligned} & y=e^{3 \sin ^{-1} x} \\ & \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}} \end{aligned} \end{aligned}$

$\sqrt{1-x^2} \frac{d y}{d x}=3 y$

Again differentiate

$\begin{aligned} & \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\ & (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right) \end{aligned}$

So value of $3 y^{\prime}\left(\sqrt{1-x^2}\right)$ at $x=\frac{1}{2}$

$\begin{aligned} & 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\ & =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}} \end{aligned}$

Given the polynomial function:

$f(x) = ax^3 + bx^2 + cx + 41$

We are provided the following conditions from the problem:

1. $f(1) = 40$

2. $f^{\prime}(1) = 2$

3. $f^{\prime \prime}(1) = 4$

First, calculate $f(1)$:

$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$

Simplifying, we get:

$a + b + c + 41 = 40$

Therefore:

$a + b + c = -1$

Next, calculate the first derivative $f^{\prime}(x)$:

$f^{\prime}(x) = 3ax^2 + 2bx + c$

Given $f^{\prime}(1) = 2$:

$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$

Simplifying, we get:

$3a + 2b + c = 2$

Next, calculate the second derivative $f^{\prime \prime}(x)$:

$f^{\prime \prime}(x) = 6ax + 2b$

Given $f^{\prime \prime}(1) = 4$:

$f^{\prime \prime}(1) = 6a(1) + 2b = 4$

Simplifying, we get:

$6a + 2b = 4$

Dividing the entire equation by 2:

$3a + b = 2$

We now have three equations:

1. $a + b + c = -1$

2. $3a + 2b + c = 2$

3. $3a + b = 2$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$(3a + 2b + c) - (3a + b) = 2 - 2$

Which simplifies to:

$b + c = 0$

So,

$c = -b$

Substitute $c = -b$ into the first equation:

$(a + b - b = -1)$

Simplifying, we get:

$a = -1$

Now substitute $a = -1$ into the third equation:

$3(-1) + b = 2$

Which simplifies to:

$-3 + b = 2$

Therefore:

$b = 5$

Next, since $c = -b$:

$c = -5$

Finally, we need to find $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$

Simplifying, we get:

$1 + 25 + 25 = 51$

Therefore, the answer is:

Option B: 51

To determine $g^{\prime}(0)$, we start by applying the chain rule and product rule to find the derivative of the given function $g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$.

The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by:

$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $

Let's denote $u(x) = h(\mathrm{e}^x)$ and $v(x) = \mathrm{e}^{h(x)}$.

First, we need to find $u'(x)$ and $v'(x)$. Using the chain rule, we find:

$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $

Now, the derivative of $v(x)$ is:

$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $

Using the product rule, we get the derivative of $g(x)$:

$ g'(x) = u'(x) v(x) + u(x) v'(x) $

Substituting $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the above expression, we get:

$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $

Next, we need to evaluate this at $x = 0$:

First, we know that:

$ h(0) = 0 $

$ h(1) = 1 $

$ h'(0) = 2 $

$ h'(1) = 2 $

Substituting $x = 0$ into the expressions, we get:

$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $

$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $

$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $

$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $

Therefore, evaluating $g'(0)$:

$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $

$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $

$ g'(0) = 2 + 2 = 4 $

Thus, the value of $g^{\prime}(0)$ is 4, which corresponds to Option A.

The correct answer is Option A: 4.

Given the function:

$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $

we need to find its second derivative at specific points.

First, let’s compute the first derivative $ f^{\prime}(x) $:

$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $

Next, the second derivative $ f^{\prime \prime}(x) $ is:

$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $

Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:

$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $

Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:

$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $

Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.

Let $f^{\prime}(1)=k$

$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$

$\begin{aligned} & \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}$

Given information is not complete.

$\begin{aligned} & y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\ & y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\ & =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \end{aligned}$

$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$

$\begin{aligned} & \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\ & \therefore \quad y+y^{\prime}+y^{\prime \prime}=2 \end{aligned}$

Given that $(g \circ f)(x) = x$ for all $x \in \mathbf{R}$. This means $g(f(x)) = x$ for all $x \in \mathbf{R}$. Differentiating both sides with respect to $x$, we get:

$g'(f(x)) \cdot f'(x) = 1$

Now, we want to find the value of $8g'(2)$. To do this, we need to find a value of $x$ such that $f(x) = 2$. Let's solve for $x$:

$x^5 + 2e^{x/4} = 2$

By inspection, we see that $x = 0$ is a solution. Therefore, $f(0) = 2$. Now, we can substitute this into our differentiated equation:

$g'(f(0)) \cdot f'(0) = 1$

$g'(2) \cdot f'(0) = 1$

Let's find $f'(0)$:

$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$

$f'(0) = \frac{1}{2}$

Substituting this back into our equation:

$g'(2) \cdot \frac{1}{2} = 1$

$g'(2) = 2$

Finally, we can calculate $8g'(2)$:

$8g'(2) = 8 \cdot 2 = \boxed{16}$

$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$

$\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned}$

Partially differentiating w. r. t. x

$\begin{aligned} & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \mathrm{y} \rightarrow \mathrm{x} \\ & \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0 \end{aligned}$

$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$

Also $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$

$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$

$\begin{aligned} & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{array}\right| \\ & \mathrm{f}(\mathrm{x})=45 \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ & \end{aligned}$

$\begin{aligned} & y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\ & \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4} \end{aligned}$

Again,

$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

Again

$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

at $\mathrm{x}=\frac{1}{2}$,

$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$

Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$

$\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\ & =\sqrt{\pi} \end{aligned}$

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.

$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $

$\therefore$ ${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$

$\therefore$ ${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$

and ${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$

$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$

$\therefore$ ${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$

Let $f(x) = {\log _{\cos x}}\cos ec\,x$

$ = {{\log \cos ec\,x} \over {\log \cos x}}$

$ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}$

at $x = {\pi \over 4}$

$f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}$

$\therefore$ ${\log _e}2f'(x)$ at $x = {\pi \over 4} = 4$

${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$

Differentiating on both side

$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$

${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $

Square on both side

${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$

Diff on both side

$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$

$xy' + y''{x^2} + 25y = 0$

$f(x) = 3{x^2} + 1$

f'(x) is bijective function

and $f(g(x)) = x \Rightarrow g(x)$ is inverse of f(x)

$g(f(x)) = x$

$g'(f(x))\,.\,f'(x) = 1$

$g'(f(x)) = {1 \over {3{x^2} + 1}}$

Put x = 4 we get

$g'(63) = {1 \over {49}}$

Let ${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$

$\therefore$ $y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$

$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$

$\therefore$ $y = {\pi \over 4} - {\theta \over 2}$

$y = {\pi \over 4} - {{{x^3}} \over 2}$

$\therefore$ $y' = {{ - 3{x^2}} \over 2}$

$y'' = - 3x$

$\therefore$ ${x^2}y'' - 6y + {{3\pi } \over 2} = 0$