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
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