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Firstly we identify the problem, in this case we have 2 equations which contain a total of 2 unknowns, it is therefore simultaneous equations.Our first step is to rearrange an equation of our choice to gi...

Begin with the equation: y = x⁴-8x²+3. Differentiate by bringing the power down and reducing the power by 1 of each of the terms with x in and constant terms (3) become zero. dy/dx =...

Set the function equal to y so y=1-2x^3now rearrange this to set x as the subject 2x^3= 1-yx^3 = (1-y)/2x = ((1-y)/2)^1/3therefore the inverse is f^-1 = ((1-x)/2)^1/3

This is done using the product rule: dy/dx=udv/dx +vdu/dxset y=uv therefore u=x^2 v=cos(x)differentiate these with respect to x du/dx= 2x as you multiply by the power and then subtract the power by 1dv/dx...

Well first, we could start with a straight line y = x. You should remember from GCSE that the equation of a straight line is given by y = mx + c, where m here is equal to 1.