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 ▸Use the double angle formulae and the identity cos(A+B)≡cos(A)cos(B)−sin(A)sin(B) to obtain an expression for cos 3x in terms of cos x only
Find the integral of xcosx(dx)
A curve is described by the equation (x^2)+4xy+(y^2)+27=0. The tangent to the point P, which lies on the curve, is parallel to the x-axis. Given the x-co-ordinate of P is negative, find the co-ordinates of P.
Given that f(x)= (3+x^2)(x^1/2-7x). Find f'(x) (5marks)