differentiation 急!!!

2013-10-21 8:36 am
f(x)=1/[(x^6)+1]

find f'(x) , f''(x) and f'''(x)

回答 (1)

2013-10-21 9:34 am
✔ 最佳答案
y = 1/(x⁶+1)

dy/dx = [d(x⁶+1)⁻¹/d(x⁶+1)][d(x⁶+1)/dx]

dy/dx = [-1/(x⁶+1)²][6x⁵]

dy/dx = -6x⁵/(x⁶+1)²

Let y = (-6x⁵)/(x⁶+1)²

dy/dx = (x⁶+1)²[d(-6x⁵)/dx] + (-6x⁵)[d(x⁶+1)⁻²/d(x⁶+1)][d(x⁶+1)/dx]

dy/dx = -30x⁴/(x⁶+1)² - (6x⁵)(-2)(x⁶+1)⁻³(6x⁵)

dy/dx =-30x⁴/(x⁶+1)² + 72x¹⁰/(x⁶+1)³

dy/dx = [-30x⁴/(x⁶+1) + 72x¹⁰]/(x⁶+1)³

dy/dx = (42x¹⁰ - 30x⁴)/(x⁶+1)³

dy/dx = 42x¹⁰/(x⁶+1)³ - 30x⁴/(x⁶+1)³

你會發覺用乘的 formula 比用除的 formula 計得輕鬆小小,

只要將 -6x⁵/(x⁶+1)² 寫做-6x⁵ 乘 (x⁶+1)⁻² 便可以用乘了。自己再試試啦!


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