Given that f(x)= (4/x) - 3x + 2 find i) f'(x) and ii) f''(1/2)

This question is asking us to find the first order derivative of f(x) with respect to x, then the second order derivative when x=1/2i) Firstly, let's consider (d/dx)4/x:There are two ways to think about differentiating a fraction like this one. One could use the quotient rule: y=u/v then dy/dx=(v(du/dx) - u(dv/dx))/v^2. However, the simplest method in this case would be to remove the fraction using indices and then use the following differentiation rule: if y=x^2 then dy/dx= 2x.Hence, in this case we would have 4/x = 4x^-1, making f(x)= 4x^-1 - 3x + 2.From here we can use the previously stated differentiation rule to find f'(x). We bring down the -1 and times it by the 4, then take away one from the power: y=4x^-1 then dy/dx = -4x^-2 .We can use the same rule for y=-3x, bring down and multiply the power (in this case 13), then take one from the power (remember anything to the power of 0 equals 1!) : y=-3x then dy/dx= -3Finally, when we differentiate any integer without an x, it differentiates to zero, giving the solution:f'(x)= -4x^-2 - 3ii) To find f''(x) we differentiate by the same method:y=-4x^-2 then dy/dy= (-4-2)x^-3 = 8x^-3Again the integer -3 differentiates to 0, meaning f''(x) = 8x^-3Finally, to find f''(1/2) we need to sub in x = 1/2:f''(1/2) = 8*(1/2)^-3 = 8/(1/2)^3 = 8/(1/8) = (8*8)/1 =64