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) \( y=\sqrt{7-3 x}, x=1 \) (c) \( y=\frac{1}{3 x-2}, y=1 \) (d) \( y=\frac{1}{(2 x-5)^{3}}, y=\frac{1}{8} \)

To find \( \frac{\mathrm{d} y}{\mathrm{~d} x} \) and the gradient of the curve at the specified points, we can differentiate the given equations and then substitute the specified values. ### (a) \( y=(3 x-2)^{3}, x=2 \) First, we differentiate: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = 3(3x - 2)^{2} \cdot 3 = 9(3x - 2)^{2} \] Now, evaluate at \( x = 2 \): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = 9(3(2) - 2)^{2} = 9(6 - 2)^{2} = 9(4)^{2} = 9 \cdot 16 = 144 \] ### (b) \( y=\sqrt{7-3 x}, x=1 \) Differentiate: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2\sqrt{7 - 3x}} \cdot (-3) = \frac{-3}{2\sqrt{7 - 3x}} \] Now, evaluate at \( x = 1 \): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3}{2\sqrt{7 - 3(1)}} = \frac{-3}{2\sqrt{4}} = \frac{-3}{4} \] ### (c) \( y=\frac{1}{3 x-2}, y=1 \) To find \( x \) when \( y = 1 \): \[ 1 = \frac{1}{3x - 2} \implies 3x - 2 = 1 \implies 3x = 3 \implies x = 1 \] Differentiate: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-1 \cdot 3}{(3x - 2)^{2}} = \frac{-3}{(3x - 2)^{2}} \] Evaluate at \( x = 1 \): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3}{(3(1) - 2)^{2}} = \frac{-3}{(1)^{2}} = -3 \] ### (d) \( y=\frac{1}{(2 x-5)^{3}}, y=\frac{1}{8} \) To find \( x \) when \( y = \frac{1}{8} \): \[ \frac{1}{8} = \frac{1}{(2x - 5)^{3}} \implies (2x - 5)^{3} = 8 \implies 2x - 5 = 2 \implies 2x = 7 \implies x = \frac{7}{2} \] Differentiate: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3(2)}{(2x - 5)^{4}} = -\frac{6}{(2x - 5)^{4}} \] Evaluate at \( x = \frac{7}{2} \): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6}{(2 \cdot \frac{7}{2} - 5)^{4}} = -\frac{6}{(7 - 5)^{4}} = -\frac{6}{2^{4}} = -\frac{6}{16} = -\frac{3}{8} \] ### Summary of Results: - (a) Gradient at \( x=2 \): 144 - (b) Gradient at \( x=1 \): \(-\frac{3}{4}\) - (c) Gradient at \( y=1 \): -3 - (d) Gradient at \( y=\frac{1}{8} \): \(-\frac{3}{8}\)