Differentiate with respect to x: i) y=x^3ln(2x) ii) y=(x+sin(2x))^3

i) There first step is to acknowledge the need for both product rule (d(uv)/dx=v.du/dx+u.dv/dx) and chain rule (dz/dx=dz/dy*dy/dx). Here, u=x^3 and v=ln(2x). Therefore, du/dx=3x^2 which is a standard differentiation. dv/dx requires the use of the chain rule: z=ln(2x), y=2x and x=x. Following the formula: dz/dx=1/(2x)2 and so dv/dx=1/x. Subbing u, v, du/dx and dv/dx into the prodcut rule formula gives the final answer: dy/dx = 3x^2ln(2x)+x^31/x = x^2(3ln(2x)+1) = x^2(ln(8x^3)+1).

ii)Only chain rule is needed to answer this part of the question (recalling dz/dx=dz/dy*dy/dx). This time z=(x+sin(2x))^3, y=x+sin(2x) and x=x. Therefore, dz/dy = d[y^3]/dy = 3y^2 and dy/dx = d[x+sin(2x)]/dx = 1+2cos(2x) remebering that d[sin(x)]/dx=cos(x). Now all the is required is to sub dz/dy and dy/dx back into the chain rule formula. The final answer is as follows: dy/dx = 3(x+sin(2x))^2(1+2cos(2x))