5 Find \( \frac{\mathrm{d} y}{\mathrm{~d} x} \) and the gradient of the curve at the point with the given value of \( x \) or \( y \). (a) \( y=(3 x-2)^{3}, x=2 \) (b) \( y=\sqrt{7-3 x}, x=1 \) (a)

To find \( \frac{\mathrm{d} y}{\mathrm{~d} x} \) for \( y = (3x - 2)^3 \), we apply the chain rule. The derivative is: \[ \frac{\mathrm{d} y}{\mathrm{~d} x} = 3(3x - 2)^2 \cdot 3 = 9(3x - 2)^2. \] Now, substituting \( x = 2 \): \[ \frac{\mathrm{d} y}{\mathrm{~d} x} = 9(3(2) - 2)^2 = 9(6 - 2)^2 = 9(4)^2 = 144. \] Thus, the gradient of the curve at \( x = 2 \) is \( 144 \). (b) For \( y = \sqrt{7 - 3x} \), we again use the chain rule to find the derivative: \[ \frac{\mathrm{d} y}{\mathrm{~d} x} = \frac{1}{2\sqrt{7 - 3x}} \cdot (-3) = \frac{-3}{2\sqrt{7 - 3x}}. \] Now, substituting \( x = 1 \): \[ \frac{\mathrm{d} y}{\mathrm{~d} x} = \frac{-3}{2\sqrt{7 - 3(1)}} = \frac{-3}{2\sqrt{7 - 3}} = \frac{-3}{2\sqrt{4}} = \frac{-3}{4}. \] Therefore, the gradient of the curve at \( x = 1 \) is \( -\frac{3}{4} \).