Search over 10,000 free study notes

We start by drawing a diagram which illustrates the question. First draw the x-y plane and the two curves curve y = x^2 and y = x^3. Notice that the two curves intersect at x=0 and x=1, and in the range ...

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.