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

Related Maths A Level answers

All answers ▸

Differentiate y=(x-1)^4 with respect to x.


Show that the equation 5sin(x) = 1 + 2 [cos(x)]^2 can be written in the form 2[sin(x)]^2 + 5 sin(x)-3=0


How do you solve the integral of ln(x)


Given that y = 5x^2 - 4/(x^3), x not equal to 0, find dy/dx.