Use logarithm properties to simplify the expression.

Here are the solutions to the questions: 1. (i) a) Differentiate ((3x^4)/( x)) with respect to x. Step 1: Use logarithm properties to simplify the expression. y = ((3x^4)/( x)) = (3x^4) - ( x) y = 3 + (x^4) - ( x) y = 3 + 4 x - ( x) Step 2: Differentiate each term with respect to x. (dy)/(dx) = (d)/(dx)( 3) + (d)/(dx)(4 x) - (d)/(dx)(( x)) (dy)/(dx) = 0 + 4((1)/(x)) - ((1)/( x) · x) (dy)/(dx) = (4)/(x) - ( x)/( x) (dy)/(dx) = (4)/(x) - x The derivative is (4)/(x) - x. 1. (i) b) A curve is defined parametrically by x = t^3, y = t(1 - (1)/(4)t^3). Find the equation of the tangent to the curve at the point where t = -1. Step 1: Find the coordinates (x_0, y_0) of the point of tangency when t = -1. x_0 = (-1)^3 = -1 y_0 = (-1)(1 - (1)/(4)(-1)^3) = (-1)(1 - (1)/(4)(-1)) = (-1)(1 + (1)/(4)) = (-1)((5)/(4)) = -(5)/(4) The point is (-1, -(5)/(4)). Step 2: Find (dx)/(dt) and (dy)/(dt). x = t^3 (dx)/(dt) = 3t^2 y = t - (1)/(4)t^4 (dy)/(dt) = 1 - (1)/(4)(4t^3) = 1 - t^3 Step 3: Find the gradient (dy)/(dx) using the chain rule (dy)/(dx) = (dy/dt)/(dx/dt). (dy)/(dx) = (1 - t^3)/(3t^2) Step 4: Evaluate the gradient at t = -1. m = .(dy)/(dx)|_t=-1 = (1 - (-1)^3)/(3(-1)^2) = (1 - (-1))/(3(1)) = (1+1)/(3) = (2)/(3) Step 5: Use the point-slope form of a line y - y_0 = m(x - x_0). y - (-(5)/(4)) = (2)/(3)(x - (-1)) y + (5)/(4) = (2)/(3)(x + 1) Step 6: Clear denominators and rearrange into general form. Multiply by 12: 12(y + (5)/(4)) = 12((2)/(3)(x + 1)) 12y + 15 = 8(x + 1) 12y + 15 = 8x + 8 8x - 12y + 8 - 15 = 0 8x - 12y - 7 = 0 The equation of the tangent is 8x - 12y - 7 = 0. 1. (ii) a) Resolve (6)/((1+2x)(3-x)) into partial fractions. Step 1: Set up the partial fraction decomposition. (6)/((1+2x)(3-x)) = (A)/(1+2x) + (B)/(3-x) Step 2: Multiply both sides by (1+2x)(3-x) to clear denominators