How do I differentiate a quadratic to the power n?

To do this we will use the chain rule, whereby dy/dx = dy/du * du/dx. So if y = (ax^2+bx+c)^n then we will say that u = ax^2+bx+c. Therefore y =u^n. So to find dy/dx we differentiate u with respect to x, which = 2ax +b, and multiply this by the differential of y =u^n, which is nu^(n-1). Therefore dy/dx = nu^(n-1) * (2ax+b) Subbing the original equation in for u leads to dy/dx = n(2ax+b)(ax^2+bx+c)^(n-1)

AA
Answered by Alex A. Maths tutor

5574 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given y = 2sin(θ) and x = 3cos(θ) find dy/dx.


The curve has equation y = x^3 - x^2 - 5x + 7 and the straight line has equation y = x + 7. One point of intersection, B, has coordinates (0, 7). Find the other two points of intersection, A and C.


How do you derive the quadratic formula?


AQA PC4 2015 Q5 // A) Find the gradient at P. B) Find the equation of the normal to the curve at P C)The normal P intersects at the curve again at the point Q(cos2q, sin q) Hence find the x-coordinate of Q.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning