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First step is to seperate the variables (EQ1) : (1/y^2) dy = 6x Then we integrate each side seperately giving us (EQ2) : -1/y = 3x^2 + C (remembering to add 1 to the power and divide by the new power) s...
I would differentiate it and then turn it into an equation to find the points where the gradient equals zero. With these points at hand, I would take a second derivative, this tells me how the gradient ch...
(a) 9x+11/(2x+3)(x-1) = A/2x+3 + B/x-1 9x+11 = (x-1)A + (2x+3)B Setting x=1 gives B=4, setting x=-3/2 gives A=1 so the final answer is 1/(2x+3) + 4/(x-1)
y = exx2 + ex : The derivative can be found using the chain rule. (A reminder of the chain rule: if y is a product of two funct...
This is a very common mistake, so don't worry. To see why this is the case, let's go back through our notes to the "laws of indices" and recall the rule for adding powers; amTDAnswered by Tutor56252 D. • Maths tutor8851 Views
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