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The product rule is used to differentiate this since we are trying to differentiate the product of 2 parts--x and ln(x)So using the product rule which is d/dx=u.(dv/dx) +v.(du/dx)let u=x and v=ln(x)then d...

Start with finding limits by setting 3x - x^2 = 0, then factorise x(3 - x) = 0. Therefore x = 0 or 3. The area is the integral of 3x - x^2 between x = 0 and 3, sub in 3 and 0 into 3(x^2)/2 - (x^3)/3, whic...

2x/[(x+1)(2x-4)] = [A/(x+1) + B/(2x-4)]x = A(x-2) + B(x+1)x = -1-1 = A(-3)A = 1/3x = 2 2 = B(3)B = 2/3 therefore.... 2x/[(x+1)(2x-4)] ...

You find the area of the circle (using radius²* π). You divide your angle by 360 (or by 2π). You multiply the 2 answers and you find the area of the sector.

You factorise it. First, you find a common factor between a, b and c. Then, you find the 2 numbers that multiplied give you c and added give you b. If a=1, then the additive inverse of those 2 numbers ar...