find h'(t) if h(t) = 7/t^1/3 - 6/t^3/5?

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2007-11-28 08:10:11 · 2 answers · asked by jiles357 1 in Science & Mathematics ➔ Mathematics

h (t) = 7 t^(- 1/3) - 6t^(-3/5)
h ` (t) = (-7/3) t^(-4/3) + (18/5)t^(-8/5)
h ` (t) = (- 7) / (3t^(4/3)) + (18) / 5t^(8/5)

2007-11-28 10:21:04 · answer #1 · answered by Como 7 · 2⤊ 1⤋

h(t) = 7/t^1/3 - 6/t^3/5

h(t) = 7 t^(-1/3) - 6 t^(-3/5)
h'(t) = 7(-1/3)t^(-4/3) - 6(-3/5)t^(-8/5)
= -7/(3t^(4/3)) + 18/(5 t^(8/5))

2007-11-28 16:27:37 · answer #2 · answered by PeterT 5 · 1⤊ 1⤋