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y = x^3 - 2x +4/x, dy/dx = 3x^2 - 2 - 4/(x^2) = 0 at the stationary points, 3x^4 - 2x^2 - 4 = 0, substitute in u for x^2: 3u^2 - 2u - 4 = 0, use the quadratic formula: u = (-(-2) +- sqrt((-2)^2 - 4 3 (-4)))/...

Use implicit differentiation 1) 8x - 3y^2 dy/dx - 4y - 4x dy/dx +2^y*ln2 * dy/dx = 0 You then sub in the points P (-2,4) 2) 8(-2) - 3(4)^2 *dy/dx - 4(4) - 4(-2) *dy/dx + 2^(4) *ln2 * dy/dx = 0 Rearrange to g...

Let a^b >b^a, then blna>alnb, (lna)/a > (lnb)/b, Thus we graph the function (lnx)/x, We can see that this tends towards 0 as x tends towards infinity. We can also see that it is increasing from x=0 ...

when integrating a function which describes a certain curve, the answer we get gives us the area under the curve. When integrating a general polynomial you simply add 1 to the power and divide by the new pow...

The first thing to note about this question is that it's a simultaneous equations question, involving a quadratic. We will have to use substution in order to solve it. So first of all we need to chose which ...