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1/y² dy/dx = sec²(x)∫ 1/y² dy/dx dx = ∫ sec²(x) dx-1/y + C₁ = tan(x) + C₂y = -1/(tan(x) + A) where A = C₂ - C₁y(0) = -1/A so y(0) = 1 means A = -1. Finished!

To help the student understand how to expand the brackets I would start by splitting the equation to:3x²(x - 3) + 6x(x - 3) + 7(x - 3)I then would ask the student to expand these brackets to:3...

To find a gradient at a given point, first we differentiate then we sub in the x coordinate of the point. To differentiate 4(8x+2)⁴ we must use the chain rule. First we let u=...

Here we have an example of a quadratic equation in terms of x. Start by manipulating the equation until it is equal to 0: subtract 2(x^2) from both sides to get (x^2)+2x-4=3x+8. Then subtract 3x from both...