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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 ...

In order to complete this question we need to use the quotient rule (i.e. if an equation is of the form h(x)=f(x)/g(x) then h'(x)=(g(x)f'(x)-g'(x)f(x))/g(x)^2).In our example f'(x)=cos(x),g'(x)=15...