Search over 10,000 free study notes

To differentiate we multiply by the power and take one off the power. d/dx(3x^2+1/x)= 6x-1/x^2 At a stationary point the gradient equals zero 6x-1/x^2=0 which rearranges to x=(1/6)^(1/3)

Turning point is when dy/dx = 0

if y= x² + 6x - 7

dy/dx = 2x + 6

at turning point: 2x + 6 = 0

therefore: 2x = - 6

x coordinate: x = - 3

substitute into...

In order to calculate this integral we must use the sustitution provided. x=2siny. Firtsly I will differentiate to find the dx component of the integral, so dx/dy=2cosy hence, dx=2cosydy. Now for the limi...

To find potential points of intersection between the line and the circle, we need to solve the equations simultaneously. So, we substitute y = x - 7 into the equation of the circle: (x + 2)² + ...

Because the value, x and its inverse, -1/x should multiply together to give -1. This is proof that two lines are perpendicular (or having the same magnitude of gradient but opposite signs +/- ).