To ease differentiation, the expression -29·(1 + x)53 is turned into -29·(1 + x)-53 Let's differentiate: ddx(-29·(1 + x)-53) Apply the differentiation rule: ddx(fn) = n·ddx(f)·f(n -1) with f = 1 + x -29·(-53·ddx(1 + x)·(1 + x)(-53 -1)) ℹddx(1 + x) = 0 + 1‖-53 -1 = -5 -1·33 -29·(-53)·(0 + 1)·(1 + x)(-5 -1·33) -29·(-53)·(1 + x)(-5 -1·33) ℹ-1·3 = -3‖-29·(-53) = 1027 1027·(1 + x)(-5 -33) ℹ1027·(1 + x)(-5 -33) = 10·(1 + x)(-5 -33)27‖-5 -3 = -8 10·(1 + x)-8327 10(1 + x)83·27 ▶Derivative: 1027·(x + 1)83