Finding a Derivative Using the Definition of a Derivative

y=(1/3x^3) - (5x) - (4/x)

Don't use the shortcut method.

By (1/3x³), do you mean (1/3) x³ or 1/(3x³)? I'll assume the latter.

f(x) = 1/(3x³) - 5x - 4/x

f'(x) = lim[h→0] (f(x+h) - f(x)) / h
. . . = lim[h→0] [(1/(3(x+h)³) - 5(x+h) - 4/(x+h)) - (1/(3x³) - 5x - 4/x)] / h
. . . = lim[h→0] (1/(3(x+h)³) - 1/(3x³) - 5(x+h) + 5x - 4/(x+h) + 4/x) / h
. . . = lim[h→0] [(1/(3(x+h)³) - 1/(3x³)) - (5(x+h) - 5x) - (4/(x+h) - 4/x)] / h
. . . = lim[h→0] [(1/(3(x+h)³) - 1/(3x³))/h - (5(x+h) - 5x)/h - (4/(x+h) - 4/x)/h]
. . . = lim[h→0] [(3x³ - 3(x+h)³)/(3x³(3(x+h)³)h) - (5x+5h-5x)/h - (4x - 4(x+h))/(hx(x+h))]
. . . = lim[h→0] [(3x³ - 3(x³+3x²h+3xh²+h³))/(3x³(3(x+h)³)h) - (5h)/h - (4x-4x-4h)/(hx(x+h))]
. . . = lim[h→0] [(3x³-3x³-9x²h-9xh²-3h³)) / (3x³(3(x+h)³)h) - (5h)/h - (-4h)/(hx(x+h))]
. . . = lim[h→0] [(-9x²h-9xh²-3h³)/(3x³(3(x+h)³)h) - (5h)/h - (-4h)/(hx(x+h))]
. . . = lim[h→0] [3h(-3x²-3xh-h²)/(3x³(3(x+h)³)h) - (5h)/h - (-4h)/(hx(x+h))]
. . . = lim[h→0] [(-3x²-3xh-h²)/(3x³(x+h)³) - 5 + 4/(x(x+h))]

. . . = (-3x²-0-0)/(3x³(x+0)³) - 5 + 4/(x(x+0))
. . . = -3x²/(3x⁶) - 5 + 4/x²
. . . = -1/x⁴ - 5 + 4/x²

Mαthmφm