How to differentiate a bracket raised to a power i.e. chain rule
Lets say the equation to be differentiated takes the following format y = (ax2+bx+c)n, to find dy/dx: (1) Make u equal the contents of the bracket, u=ax2+bx+c. (2) Substitute the contents of the bracket with u, y=un. (3) Differentiate y with respect to u, dy/du=nu(n-1). (4) Differentiate u with respect to x, du/dx=2ax+b. (5) Because dy/dx=(dy/du)(du/dx), we can derive. (6) dy/dx=(nu(n-1))(2ax+b). (7) Finally, substitute u with ax2+bx+c, dy/dx=(n(ax2+bx+c)(n-1))(2ax+b).
DF
