Explanation

y = x\(^2\)(1 + x)\(^{\frac{3}{2}}\)

Let u = x\(^2\) 

v = (1 + x)\(^{\frac{3}{2}}\)

\(\frac{du}{dx} 2x\)

\(\frac{dv}{dx} = \frac{3}{2}\)(1 + x)\(^\frac{1}{2}\)

\(\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}\) 

= (1 + x)\(^{\frac{3}{2}}\)(2x) + x\(^2\)\(\frac{3}{2}(1 + x)^{\frac{1}{2}}\)

= 2x(1 + x)\(^{\frac{3}{2}}\) + \(\frac{3x^2(1 + x)^{\frac{1}{2}}}{2}\)

(b) 

Let the coordinates of the centre of a circle be C(h,k) which lies on the line

2y - x = 3 , therefore 2k — h = 3. ………………..(1).

The circle passes through the points P(2,3) and Q(6,7) ,then :—

CP = CQ = radius of the circle.

or, (CP)\(^2\) = (CQ)\(^2\).

or, (h - 2)\(^2\) + (k-3)\(^2\) = (h  - 6)\(^2\) + (k - 7)\(^2\).

or, h\(^2\) – 4h + 4 + k\(^2\) – 6k + 9 = h\(^2\) – 12h + 36 + k\(^2\) – 14k + 49.

or, 8h + 8k = 85 - 13 = 72.

or, h + k = 9…………………(2) , from eqn.(1) and (2).

h = 5, k = 4. Thus the coordinates of the centre of a circle are (5,4).

Radius (r) = CP = \(\sqrt{(5–2)^2 + (4 – 3)^2}\) = \(\sqrt{10}\) units.

Equation of a circle is:—

(x - 5)\(^2\) + (y - 4)\(^2\) = \(\sqrt{(10)}^2\)

x\(^2\) - 10x + 25 + y\(^2\) - 8y + 16 = 10

x\(^2\) + y\(^2\) - 10x - 8y + 25 + 16 - 10 = 0

x\(^2\) + y\(^2\) - 10x - 8y + 31 = 0

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