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The first step is to split the fraction into 2 separate fractions using partial fractions techniques. Write 3x+3/2x^2+3x as A/x + B/2x+3 and solve to get A = 1, B = 1. We have now converted 3x+3/2x^2+3x i...

In order to find the points where these functions meet, we can equate them to get 3x-2 = x^2+4x-8 .

Subtraction (3x-2) from both sides, we get x^2+x-6 =0 which we can factorise to get (x+3)(x-2)=0<...

We have a "fraction" which we wish to differentiate, so we use the quotient rule with u=sin(x) and v=cos(x).

This means that d/dx of u/v = (vdu/dx - udv/dx)/(v^2).

y = cos(x)/sec²(x) = cos³(x)

y = cos(x)(1-sin²(x)) = cos(x) - cos(x) sin²(x)

-> sin(x) - sin³(x)/3 + c

First let a = b = x such that:          

          cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

becomes:

          cos(x + x) = cos(x)cos(x) - sin(x)sin(x)

Leading to:

     ...