(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 pa...

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(a) To find the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\), we can use the product rule of differentiation, which states that if we have two functions u(x) and v(x), then the derivative of their product is given by: \((u\cdot v)' = u'v + uv'\) where u' and v' denote the derivatives of u and v with respect to x, respectively. Let u(x) = x\(^2\) and v(x) = (1 + x)\(^{\frac{3}{2}}\). Then, we have: u'(x) = 2x (by the power rule of differentiation) v'(x) = \(\frac{3}{2}\)(1 + x)\(\frac{1}{2}\) = \(\frac{3}{2}\)\(\sqrt{1 + x}\) (by the chain rule of differentiation) Using the product rule, we can now find the derivative of y as follows: y' = u'v + uv' = 2x(1 + x)\(^{\frac{3}{2}}\) + x\(^2\)\(\frac{3}{2}\)\(\sqrt{1 + x}\) = x(4 + 3x\(\sqrt{1 + x}\)). Therefore, the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\) with respect to x is x(4 + 3x\(\sqrt{1 + x}\)). (b) Since the centre of the circle lies on the line 2y - x = 3, we can express the centre as a point (h, k) such that 2k - h = 3. Since the circle passes through points P(2, 3) and Q(6, 7), we can use the distance formula to write two equations involving h and k, as follows: \((h-2)^2 + (k-3)^2 = r^2\) (from P) \((h-6)^2 + (k-7)^2 = r^2\) (from Q) where r is the radius of the circle. Expanding both equations and subtracting them, we get: \(8h + 8k = 32\) Substituting k = \(\frac{h+3}{2}\) (from the line equation 2k - h = 3) into the above equation, we get: \(h^2 - 14h + 25 = 0\) Solving this quadratic equation, we get h = 5 or h = 2.5. If h = 5, then from the line equation 2k - h = 3, we get k = 4. If h = 2.5, then k = 2. Thus, the centre of the circle is either (5, 4) or (2.5, 2). To find the radius of the circle, we can use either of the two equations we obtained earlier (from P or Q). Let's use the equation from P: \((h-2)^2 + (k-3)^2 = r^2\) Substituting (h, k) = (5, 4), we get: \((5-2)^2 + (4-3)^2 = r^2\) Simplifying, we get: \(r^2 = 10\) Taking the square root of both sides, we get: \(r = \sqrt{10}\) Therefore, the equation