Higher order derivatives Questions in English

(C) Given the function: $y = ae^x + be^{-x} + c$
Differentiating with respect to $x$ once:
$y^{\prime} = \frac{d}{dx}(ae^x + be^{-x} + c) = ae^x - be^{-x}$
Differentiating again with respect to $x$:
$y^{\prime \prime} = \frac{d}{dx}(ae^x - be^{-x}) = ae^x + be^{-x}$
Differentiating a third time with respect to $x$:
$y^{\prime \prime \prime} = \frac{d}{dx}(ae^x + be^{-x}) = ae^x - be^{-x}$
Comparing this result with the first derivative,we see that $y^{\prime \prime \prime} = y^{\prime}$.

(B) Given $y=a \cos (\log x)+b \sin (\log x)$.
First,differentiate with respect to $x$:
$y^{\prime} = \frac{d}{dx} [a \cos (\log x)+b \sin (\log x)] = -a \sin (\log x) \cdot \frac{1}{x} + b \cos (\log x) \cdot \frac{1}{x} = \frac{-a \sin (\log x) + b \cos (\log x)}{x}$.
So,$x y^{\prime} = -a \sin (\log x) + b \cos (\log x)$.
Now,differentiate again with respect to $x$:
$\frac{d}{dx} (x y^{\prime}) = \frac{d}{dx} [-a \sin (\log x) + b \cos (\log x)]$.
Using the product rule on the left side: $x y^{\prime \prime} + y^{\prime} = -a \cos (\log x) \cdot \frac{1}{x} - b \sin (\log x) \cdot \frac{1}{x}$.
Multiply both sides by $x$:
$x^2 y^{\prime \prime} + x y^{\prime} = -[a \cos (\log x) + b \sin (\log x)]$.
Since $y = a \cos (\log x) + b \sin (\log x)$,we have:
$x^2 y^{\prime \prime} + x y^{\prime} = -y$.

(C) Given that,$f(x)=\frac{x-1}{e^x} \dots (i)$
Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{e^x(1) - (x-1)e^x}{(e^x)^2} = \frac{e^x(1 - x + 1)}{e^{2x}} = \frac{2-x}{e^x} \dots (ii)$
Substituting $x=0$ in Eq. $(ii)$:
$f'(0) = \frac{2-0}{e^0} = 2 \dots (iii)$
Now,differentiate Eq. $(ii)$ with respect to $x$:
$f''(x) = \frac{e^x(-1) - (2-x)e^x}{(e^x)^2} = \frac{e^x(-1 - 2 + x)}{e^{2x}} = \frac{x-3}{e^x} \dots (iv)$
Substituting $x=0$ in Eq. $(iv)$:
$f''(0) = \frac{0-3}{e^0} = -3 \dots (v)$
Adding Eq. $(iii)$ and Eq. $(v)$:
$f'(0) + f''(0) = 2 + (-3) = -1$

(A) Given $y = \sin ax + \cos bx$.
First,differentiate with respect to $x$:
$y' = \frac{d}{dx}(\sin ax) + \frac{d}{dx}(\cos bx) = a \cos ax - b \sin bx$.
Next,differentiate again to find the second derivative:
$y'' = \frac{d}{dx}(a \cos ax - b \sin bx) = -a^2 \sin ax - b^2 \cos bx$.
Now,substitute $y''$ and $y$ into the expression $y'' + b^2 y$:
$y'' + b^2 y = (-a^2 \sin ax - b^2 \cos bx) + b^2(\sin ax + \cos bx)$.
$y'' + b^2 y = -a^2 \sin ax - b^2 \cos bx + b^2 \sin ax + b^2 \cos bx$.
$y'' + b^2 y = (b^2 - a^2) \sin ax$.

(B) Given $y=e^{ax}(\cos bx+\sin bx)$.
First derivative: $\frac{dy}{dx}=ae^{ax}(\cos bx+\sin bx)+e^{ax}(-b\sin bx+b\cos bx) = ay+be^{ax}(\cos bx-\sin bx)$.
Rearranging gives $be^{ax}(\cos bx-\sin bx) = \frac{dy}{dx}-ay$.
Second derivative: $\frac{d^2y}{dx^2} = a\frac{dy}{dx} + b[ae^{ax}(\cos bx-\sin bx) + e^{ax}(-b\sin bx-b\cos bx)]$.
$\frac{d^2y}{dx^2} = a\frac{dy}{dx} + a[be^{ax}(\cos bx-\sin bx)] - b^2e^{ax}(\sin bx+\cos bx)$.
Substituting $be^{ax}(\cos bx-\sin bx) = \frac{dy}{dx}-ay$ and $e^{ax}(\cos bx+\sin bx) = y$:
$\frac{d^2y}{dx^2} = a\frac{dy}{dx} + a(\frac{dy}{dx}-ay) - b^2y$.
$\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y = 0$.
Comparing with $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$,we get $K=2a$ and $L=a^2+b^2$.
Thus,$L+bK = a^2+b^2+b(2a) = a^2+b^2+2ab = (a+b)^2$.

(A) Given: $\cos ^{-1}\left(\frac{y}{b}\right) = n \log \left(\frac{x}{n}\right)$.
Taking the derivative with respect to $x$:
$-\frac{1}{\sqrt{1 - (y/b)^2}} \cdot \frac{y_1}{b} = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.
$-\frac{1}{\sqrt{(b^2 - y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$ $\Rightarrow -\frac{b}{\sqrt{b^2 - y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$.
$-\frac{y_1}{\sqrt{b^2 - y^2}} = \frac{n}{x} \Rightarrow -x y_1 = n \sqrt{b^2 - y^2}$.
Squaring both sides: $x^2 y_1^2 = n^2 (b^2 - y^2)$.
Differentiating again with respect to $x$:
$x^2 (2 y_1 y_2) + 2x y_1^2 = n^2 (-2 y y_1)$.
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$x^2 y_2 + x y_1 = -n^2 y \Rightarrow x^2 y_2 + x y_1 + n^2 y = 0$.

(C) Let $u = \log_e x$. Then the expression becomes:
$y = \tan^{-1}\left[\frac{\log_e e - \log_e x^2}{\log_e e + \log_e x^2}\right] + \tan^{-1}\left[\frac{3+2u}{1-6u}\right]$
$y = \tan^{-1}\left[\frac{1-2u}{1+2u}\right] + \tan^{-1}\left[\frac{3+2u}{1-3(2u)}\right]$
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$y = (\tan^{-1} 1 - \tan^{-1}(2u)) + (\tan^{-1} 3 + \tan^{-1}(2u))$
$y = \tan^{-1} 1 + \tan^{-1} 3$
Since $y$ is a constant,the derivative $\frac{dy}{dx} = 0$.
Therefore,the second derivative $\frac{d^2 y}{d x^2} = 0$.

(B) Given $\cos ^{-1}\left(\frac{y}{b}\right)=n \log _e\left(\frac{x}{n}\right)$.
Differentiating with respect to $x$:
$-\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} y_1 = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.
$-\frac{y_1}{\sqrt{b^2-y^2}} = \frac{n}{x} \implies x y_1 = -n \sqrt{b^2-y^2}$.
Squaring both sides: $x^2 y_1^2 = n^2 (b^2-y^2)$.
Differentiating again with respect to $x$:
$2 x y_1^2 + x^2 \cdot 2 y_1 y_2 = -n^2 \cdot 2 y y_1$.
Dividing by $2 x y_1$ (assuming $x \neq 0, y_1 \neq 0$):
$y_1 + x y_2 = -\frac{n^2 y}{x}$.
$x y_1 + x^2 y_2 + n^2 y = 0$.
Comparing with $A y_2 + B y_1 + C y = 0$,we get $A=x^2, B=x, C=n^2$.

(A) Given the equation: $y = A x^{-1} + B x^2$.
First,differentiate $y$ with respect to $x$:
$\frac{d y}{d x} = -A x^{-2} + 2 B x$.
Next,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2 y}{d x^2} = 2 A x^{-3} + 2 B$.
Now,multiply by $x^2$:
$x^2 \frac{d^2 y}{d x^2} = x^2 (2 A x^{-3} + 2 B) = 2 A x^{-1} + 2 B x^2$.
Factor out $2$:
$x^2 \frac{d^2 y}{d x^2} = 2 (A x^{-1} + B x^2)$.
Since $y = A x^{-1} + B x^2$,we substitute $y$ back into the expression:
$x^2 \frac{d^2 y}{d x^2} = 2 y$.

(B) Given $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right)$.
Let $x = \cos \theta$,then $\sqrt{1-x^2} = \sin \theta$.
Also,let $\sin \alpha = \frac{5}{13}$,then $\cos \alpha = \frac{12}{13}$.
Substituting these into the expression for $y$:
$y = \sin^{-1}(\sin \alpha \cos \theta + \cos \alpha \sin \theta)$
$y = \sin^{-1}(\sin(\alpha + \theta)) = \alpha + \theta$
$y = \sin^{-1}(\frac{5}{13}) + \cos^{-1}(x)$
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = 0 - \frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sqrt{1-x^2}}$
Multiplying by $\sqrt{1-x^2}$ gives $y_1 \sqrt{1-x^2} = -1$.
Squaring both sides: $y_1^2 (1-x^2) = 1$.
Differentiating again with respect to $x$:
$2y_1 y_2 (1-x^2) + y_1^2 (-2x) = 0$
Dividing by $2y_1$ (assuming $y_1 \neq 0$):
$y_2(1-x^2) - x y_1 = 0$.
Comparing this with $a(1-x^2)y_2 + bxy_1 = 0$,we get $a=1$ and $b=-1$.
Thus,$(a, b) = (1, -1)$.

(B) Given $f(x) = \tan^{-1} x$.
First derivative: $f'(x) = \frac{1}{1+x^2}$.
Second derivative: $f''(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = \frac{-2x}{(1+x^2)^2}$.
We are given $f'(x) + f''(x) = 0$.
Substituting the derivatives: $\frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} = 0$.
Multiply by $(1+x^2)^2$: $(1+x^2) - 2x = 0$.
This simplifies to $x^2 - 2x + 1 = 0$,which is $(x-1)^2 = 0$.
Therefore,$x = 1$.

(C) Given $y = e^{kx}$.
Then $\frac{dy}{dx} = ke^{kx} = ky$ and $\frac{d^2y}{dx^2} = k^2e^{kx} = k^2y$.
Substituting these into the given equation:
$(\frac{d^2y}{dx^2} + \frac{dy}{dx})(\frac{dy}{dx} - y) = y\frac{dy}{dx}$
$(k^2y + ky)(ky - y) = y(ky)$
$ky(k+1) \cdot y(k-1) = ky^2$
$k(k^2 - 1)y^2 = ky^2$
Since $y = e^{kx} \neq 0$,we can divide by $y^2$:
$k(k^2 - 1) = k$
$k^3 - k = k$
$k^3 - 2k = 0$
$k(k^2 - 2) = 0$
Thus,$k = 0$ or $k^2 = 2$,which gives $k = 0, \sqrt{2}, -\sqrt{2}$.
There are three distinct values of $k$.

(A) Given $y=e^{\tan ^{-1} x}$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = e^{\tan ^{-1} x} \times \frac{d}{dx}(\tan ^{-1} x) = e^{\tan ^{-1} x} \times \frac{1}{1+x^2}$.
Since $y = e^{\tan ^{-1} x}$,we can write:
$\frac{dy}{dx} = \frac{y}{1+x^2}$.
Multiplying both sides by $(1+x^2)$,we get:
$(1+x^2) y_1 = y$.
Differentiating again with respect to $x$ using the product rule on the left side:
$\frac{d}{dx}[(1+x^2) y_1] = \frac{d}{dx}(y)$.
$(1+x^2) y_2 + y_1(2x) = y_1$.
Rearranging the terms:
$(1+x^2) y_2 + (2x - 1) y_1 = 0$.

(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.
Using partial fractions,we write:
$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^{2}}$.
Solving for coefficients,we get $A=4, B=-3, C=1$.
So,$y = 4(x+2)^{-1} - 3(x+1)^{-1} + (x+1)^{-2}$.
Differentiating with respect to $x$ once:
$y' = -4(x+2)^{-2} + 3(x+1)^{-2} - 2(x+1)^{-3}$.
Differentiating again:
$y'' = 8(x+2)^{-3} - 6(x+1)^{-3} + 6(x+1)^{-4}$.
$y'' = 2\left[\frac{4}{(x+2)^{3}} - \frac{3}{(x+1)^{3}} + \frac{3}{(x+1)^{4}}\right]$.
This matches option $A$.

(A) Given $y=e^{m \sin^{-1} x}$.
Differentiating with respect to $x$,we get $\frac{d y}{d x}=e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1-x^{2}}}$.
This implies $\sqrt{1-x^{2}} \frac{d y}{d x}=m e^{m \sin^{-1} x} = m y$.
Squaring both sides or differentiating again with respect to $x$ using the product rule:
$\frac{d}{d x} \left( \sqrt{1-x^{2}} \frac{d y}{d x} \right) = \frac{d}{d x} (m y)$.
$\sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1-x^{2}}} (-2 x) \frac{d y}{d x} = m \frac{d y}{d x}$.
Multiplying throughout by $\sqrt{1-x^{2}}$:
$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1-x^{2}} \frac{d y}{d x}$.
Substituting $\sqrt{1-x^{2}} \frac{d y}{d x} = m y$ into the equation:
$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m(m y) = m^{2} y$.
Rearranging gives $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$.
Comparing this with the given equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$,we find $k = m^{2}$.

(D) Given,$f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$
Using the properties of logarithms,$\log(e/x^2) = \log e - 2 \log x = 1 - 2 \log x$ and $\log(ex^2) = \log e + 2 \log x = 1 + 2 \log x$.
So,$f(x) = \tan^{-1}\left[\frac{1 - 2 \log x}{1 + 2 \log x}\right] + \tan^{-1}\left[\frac{3 + 2 \log x}{1 - 3(2 \log x)}\right]$.
Let $2 \log x = u$. Then $f(x) = \tan^{-1}\left[\frac{1 - u}{1 + u}\right] + \tan^{-1}\left[\frac{3 + u}{1 - 3u}\right]$.
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$f(x) = (\tan^{-1} 1 - \tan^{-1} u) + (\tan^{-1} 3 + \tan^{-1} u)$.
$f(x) = \frac{\pi}{4} + \tan^{-1} 3$.
Since $f(x)$ is a constant,its derivative $f'(x) = 0$ and the second derivative $f''(x) = 0$.

(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.
Integrating both sides with respect to $x$,we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.
Substituting $x=1$,we have $f^{\prime}(1)=g^{\prime}(1)+C_1$.
Given $f^{\prime}(1)=4$ and $g^{\prime}(1)=6$,so $4=6+C_1$,which implies $C_1=-2$.
Thus,$f^{\prime}(x)=g^{\prime}(x)-2$.
Integrating again with respect to $x$,we get $f(x)=g(x)-2x+C_2$.
Substituting $x=2$,we have $f(2)=g(2)-2(2)+C_2$.
Given $f(2)=3$ and $g(2)=9$,so $3=9-4+C_2$,which implies $3=5+C_2$,so $C_2=-2$.
Thus,$f(x)=g(x)-2x-2$.
Rearranging the terms,we get $f(x)-g(x)=-2x-2$.
For $x=1$,$f(1)-g(1)=-2(1)-2=-4$.

(C) Given $x=\sin \theta$ and $y=\sin(k \theta)$.
First,find $y_1 = \frac{dy}{dx}$:
$\frac{dy}{d\theta} = k \cos(k \theta)$ and $\frac{dx}{d\theta} = \cos \theta$.
$y_1 = \frac{dy/d\theta}{dx/d\theta} = \frac{k \cos(k \theta)}{\cos \theta} \implies y_1 \cos \theta = k \cos(k \theta)$.
Differentiating both sides with respect to $x$:
$y_2 \cos \theta - y_1 \sin \theta \cdot \frac{d\theta}{dx} = -k^2 \sin(k \theta) \cdot \frac{d\theta}{dx}$.
Since $\frac{d\theta}{dx} = \frac{1}{\cos \theta}$,we have:
$y_2 \cos \theta - y_1 \sin \theta \cdot \frac{1}{\cos \theta} = -k^2 \sin(k \theta) \cdot \frac{1}{\cos \theta}$.
Multiplying by $\cos \theta$:
$y_2 \cos^2 \theta - y_1 \sin \theta = -k^2 \sin(k \theta)$.
Since $1-x^2 = 1-\sin^2 \theta = \cos^2 \theta$ and $x = \sin \theta$:
$(1-x^2) y_2 - x y_1 = -k^2 y$.
Comparing this with $(1-x^2) y_2 - x y_1 - \alpha y = 0$,we get $\alpha y = -k^2 y$,so $\alpha = -k^2$.

(B) Given the equation $x^2+y^2=1$.
Differentiating both sides with respect to $x$,we get:
$2x + 2y y^{\prime} = 0$
Dividing by $2$,we have $x + y y^{\prime} = 0$.
Differentiating again with respect to $x$ using the product rule:
$1 + y y^{\prime \prime} + (y^{\prime}) \cdot y^{\prime} = 0$
$1 + y y^{\prime \prime} + (y^{\prime})^2 = 0$
Thus,$y y^{\prime \prime} + (y^{\prime})^2 + 1 = 0$.

(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$.
By the Fundamental Theorem of Calculus,differentiating both sides with respect to $y$,we get $\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^2}}$.
Taking the reciprocal,we have $\frac{dy}{dx} = \sqrt{1 + 9y^2}$.
Differentiating both sides with respect to $x$ using the chain rule,we get $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sqrt{1 + 9y^2}) = \frac{1}{2\sqrt{1 + 9y^2}} \cdot (18y) \cdot \frac{dy}{dx}$.
Substituting $\frac{dy}{dx} = \sqrt{1 + 9y^2}$,we get $\frac{d^2y}{dx^2} = \frac{9y}{\sqrt{1 + 9y^2}} \cdot \sqrt{1 + 9y^2} = 9y$.
Comparing $\frac{d^2y}{dx^2} = 9y$ with $\frac{d^2y}{dx^2} = ay$,we find $a = 9$.

(B) Given $U(x) = \frac{Lx+M}{x^2-2Bx+C}$.
Rearranging,we get $U(x)(x^2-2Bx+C) = Lx+M$.
Differentiating both sides with respect to $x$ using the product rule:
$U_1(x^2-2Bx+C) + U(2x-2B) = L$.
Differentiating again with respect to $x$:
$U_2(x^2-2Bx+C) + U_1(2x-2B) + U_1(2x-2B) + U(2) = 0$.
Simplifying the expression:
$U_2(x^2-2Bx+C) + U_1(4x-4B) + 2U = 0$.
Comparing this with $PU_2 + QU_1 + RU = 0$,we get:
$P = x^2-2Bx+C$,$Q = 4(x-B)$,and $R = 2$.

(D) Given $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$.
First derivative: $f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2f^{\prime\prime}(2)$.
Second derivative: $f^{\prime\prime}(x) = 6x + 2f^{\prime}(1)$.
Third derivative: $f^{\prime\prime\prime}(x) = 6$.
Now,evaluate the constants:
$f^{\prime\prime}(2) = 6(2) + 2f^{\prime}(1) = 12 + 2f^{\prime}(1)$.
$f^{\prime\prime\prime}(3) = 6$.
Substitute these into the expression for $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2(12 + 2f^{\prime}(1)) = 3x^{2} + 2x f^{\prime}(1) + 24 + 4f^{\prime}(1)$.
Set $x = 1$ to find $f^{\prime}(1)$:
$f^{\prime}(1) = 3(1)^{2} + 2(1)f^{\prime}(1) + 24 + 4f^{\prime}(1)$.
$f^{\prime}(1) = 3 + 2f^{\prime}(1) + 24 + 4f^{\prime}(1) = 27 + 6f^{\prime}(1)$.
$-5f^{\prime}(1) = 27 \implies f^{\prime}(1) = -\frac{27}{5}$.
Now substitute $f^{\prime}(1)$ back into $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^{2} + 2x(-\frac{27}{5}) + 24 + 4(-\frac{27}{5}) = 3x^{2} - \frac{54}{5}x + 24 - \frac{108}{5} = 3x^{2} - \frac{54}{5}x + \frac{12}{5}$.
Finally,calculate $f^{\prime}(5)$:
$f^{\prime}(5) = 3(5)^{2} - \frac{54}{5}(5) + \frac{12}{5} = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105 + 12}{5} = \frac{117}{5}$.

(D) Given $e^y(x+1) = 1$.
Taking the natural logarithm on both sides,we get: $y + \ln(x+1) = 0 \implies y = -\ln(x+1)$.
Differentiating with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x+1} = -(x+1)^{-1}$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2} = \left(\frac{1}{x+1}\right)^2$.
Now,calculating the expression: $\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{x+1}\right)^2 - \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2} - \frac{1}{(x+1)^2} = 0$.