Work out the gradient of the curve y=x^3(x-3) at the point (3,17)

First simplify the equation of the curve y= x^4 - 3x^3 .The gradient is the differential.To differentiate, bring down the power and take one from it.x^4 becomes 4x^3-3x^3 becomes (-3x3)= -9x^2dy/dx= 4x^3 - 9x^2Coordinates are written in (x,y) form. Hence x=3.Gradient at x=3 = 4x^3 - 9x^2 = 4(3^3) - 9(3^2) = 108 - 81 = 27

SM

Related Further Mathematics GCSE answers

All answers ▸

How can a system of two linear equations be solved?


How would I solve the following equation d^2x/dt^2 + 5dx/dt + 6x = 0


How can I show that the lines between sets of points are perpendicular?


Find the coordinates of the stationary points on the curve y=x^5 -15x^3