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In this exercise I have to find the coordinates of the vertex of the parabola. Given the general equation y= ax^2 + bx + c , the value of a is 2, the value of b is -4 and the value of c is 1.

In or...

As B is a stationary point, the value of dy/dx at this point must be equal to 0. Differentiating y gives this to be dy/dx = 6x²-2ax+8. At point B, <...

The easiest method to use in this incidence is integratation by parts.

So let u=x and dv/dx=exp(x). Therefore du/dx=1 and v=exp(x).

Then we use the formula where integral(udu/dx)=uv...

As this is a product of two functions it is necessary to use the product rule for differentiation. Therefore one of the functions must labeled v and the other u. i.e. u=x^3 and v=(x^2+1). It is then neces...

 y = ln(2x+3 / 7x^3 +1)

d/dx(2x+3 / 7x^3 + 1) by quotient rule which is(v.du/dx - u.dv/dx) / v^2  where u=2x+3 and v=7x^3 +1   gives (-27x^3 -63x^2 +2) / (7x^3 +1)^2

so d/dx(ln(2x+3 / 7x^3 +...