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Firstly you would start by differentiating the function and equating it to zero as the gradient of the function at the maximum point is zero. to differentiate this function you would use the chain rule si...

Start by differentiating the function to find points where the gradient is 0. so dy/dx = 15x^2 + 18x + 3 We can use the equation for finding the roots of a quadratic here, set a=15, b=18 and c=3 and proc...

We have to use the chain rule here. If we set u to the inside of the bracket, u = x^2 + 3 and differentiating we get du/dx = 2x. Now the original expression becomes y = u^2. Differentiating this with resp...

tanx = sinx/cosx. Express this as: sinx*(cosx)^-1. Remembering the product rule: "y = f(x)g(x), dy/dx = f'(x)g(x) + f(x)g'(x)". sinx differentiates to cosx and cosx differentiates to -sinx. Also...

This is a handy trick for quadratic equations ax^2 + bx + c = 0.

e.g. (x^2 + 5x + 6). So a = 1, b = 5 and c = 6.

To complete the square, let x^2 + 5x + 6 = 0. Then, take 6 to the other side ...