find the gradient of y=x3 X0=5
dx/dy = 3(x)2substitute x = 5so (5) squared 253 times 25 = 75
JA
Related Maths A Level answers
All answers ▸A curve, C, has equation y =(2x-3)^5. A point, P, lies on C at (w,-32). Find the value of w and the equation of the tangent of C at point, P in the form y =mx+c.
given that y = 1 when x = π, find y in terms of x for the differential equation, dy/dx = xycos(x)
How do I calculate the rate of change of something for which I don't have an equation?
Differentiate x^5 + 3x^2 - 17 with respect to x