(a) Find the derivative of y = x\(^2\) (1 +... | WAEC SSCE - Further Mathematics - 2020

Question 1 Report

(a) Find the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\) with respect to x.

(b) The centre of a circle lies on the line 2y - x = 3. If the circle passes through P(2,3) and Q(6,7), find its equation.

Answer

(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:

\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]

Factor out \((1 + x)^{1/2}\):

\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\]

(b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):

\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]

Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):

\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]

Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).

\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]

Read lesson note on Differentiation (WAEC)

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