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To solve this inequality, we need to take steps to collect like terms on each side of the inequality. To start with, we can try getting all the numbers on one side of the inequality - on the left hand sid...

You need to remember some ‘rules’:if it has the ‘co-‘ prefix, the derivative will be negative (e.g. cos x -> -sin x)if it has the ‘-sec’ suffix, the derivative will only include itself (e.g. cosec x -&...

\frac{1}{x^2+x}=\frac{A}{x} + \frac{B}{x+1}, 1=A(x+1)+Bx, let x=-1: 1=-B, B=-1let x=0: 1=A, A=1Hence, \int_{1}^{2}{\frac{1}{x^2+x}dx} = \int_{1}^{2}{\frac{1}{x}dx} - \int_{1}^{2}{\frac{1}{x+1}dx}=[ln(x)-l...

The essay is usually either a discursive or an argumentative essay. Because of this, it is best to be prepared for both possibilities; however, the discursive essay is arguably more difficult since you ha...

Using the product rule :u = x, so du/dx = 1
v = sin(x), so dv/dx = cos(x)
Therefore dy/dx = v(du/dx) + u(dv/dx) So dy/dx = sin(x) + x.cos(x)