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Please see this document which contains a scanned copy of my answer.https://docs.google.com/document/d/1A9idRZsUAy5QMLSJNDQEun-l7mSEjpRLx_Jwx44JoqY/edit?usp=sharing

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...

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)

When differentiating an expression of x and y we can use implicit differentiation when terms contain the y variable, recall that this is essentially differentiating the y term by y and then multiplying by...