(a) Differentiate \(\frac{x^{2} + 1}{(x + 1)^{2}}\) with respect to x. (b)(i) Evaluate \(\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix...

Answer Details

(a) Differentiate \( y = \dfrac{x^2 + 1}{(x+1)^2} \) using the quotient rule with \( u = x^2+1,\ v = (x+1)^2 \):

\( u' = 2x,\ v' = 2(x+1) \).

\[ \frac{dy}{dx} = \frac{2x(x+1)^2 - (x^2+1)\,2(x+1)}{(x+1)^4} = \frac{2(x+1)\big[x(x+1) - (x^2+1)\big]}{(x+1)^4}. \]

\[ = \frac{2\big[x^2 + x - x^2 - 1\big]}{(x+1)^3} = \frac{2(x-1)}{(x+1)^3}. \]

(b)(i) Expand the determinant along the first row:

\[ \Delta = 1(3\cdot3 - (-1)\cdot1) - 2(2\cdot3 - (-1)(-1)) + (-1)(2\cdot1 - 3(-1)). \]

\[ \Delta = 1(10) - 2(5) - 1(5) = 10 - 10 - 5 = -5. \]

(ii) The system \( x+2y-z=4,\ 2x+3y-z=2,\ -x+y+3z=-1 \) has coefficient determinant \(\Delta = -5 \neq 0\), so use Cramer's rule.

\( \Delta_x = \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} = 40 - 10 - 5 = 25 \Rightarrow x = \dfrac{25}{-5} = -5. \)

\( \Delta_y = \begin{vmatrix} 1 & 4 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 3 \end{vmatrix} = 5 - 20 - 0 = -15 \Rightarrow y = \dfrac{-15}{-5} = 3. \)

\( \Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 2 \\ -1 & 1 & -1 \end{vmatrix} = -5 - 0 + 20 = 15 \Rightarrow z = \dfrac{15}{-5} = -3. \)

Solution: \( x = -5,\ y = 3,\ z = -3 \).