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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...
A stationary point is a point on the function where the gradient is zero. The phrase 'stationary point' coming up in a question always indicates that differentiation may be useful to solve it. In this cas...
We use differentiation to generally find the rate of change for a function. This could also be interpreted as finding the gradient of a curve. e.g. y = x ^ 2.
If you consider the curve, it has a di...
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