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To work out the gradient of a function f(x), we need to differentiate it with respect to x, to give us f'(x). If x = a at a point, then the gradient of f(x) at that point is f'(a) (substitute a in plac...

Let us denote sin(nx) = u(x), where u is a function of x. The equation is now therefore f(x) =(u(x))^n.

For simplicity, we will write that as f(x) = u^n

By the chain rule, w...

First simplify the expression; 3x^(2)-3 to get;

[(x+1)/3(x^(2)-1)] - [1/(3x+1)] 

Using the fact that x^(2)-1 is the difference of two squares, we can simplify it to;

This is a standard parametric equations question, it is important that the methodology behind answering this question is understood.

For part a) we simply use the chain rule (dy/dx = dy/d...

Firstly, find the values of x where f'(x) = 0

f'(x) = 6x² - 54

6x² - 54 = 0

6(x+3)(x-3) = 0

x = 3, y = -108 and x = -3, y =...