f(x)=(x^1/2-1)(x^1/2-2)……(x^1/2-2008),求f'(1).用求导法则求.
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![f(x)=(x^1/2-1)(x^1/2-2)……(x^1/2-2008),求f'(1).用求导法则求.](/uploads/image/z/12476090-2-0.jpg?t=f%28x%29%3D%28x%5E1%2F2-1%29%28x%5E1%2F2-2%29%E2%80%A6%E2%80%A6%28x%5E1%2F2-2008%29%2C%E6%B1%82f%27%281%29.%E7%94%A8%E6%B1%82%E5%AF%BC%E6%B3%95%E5%88%99%E6%B1%82.)
f(x)=(x^1/2-1)(x^1/2-2)……(x^1/2-2008),求f'(1).用求导法则求.
f(x)=(x^1/2-1)(x^1/2-2)……(x^1/2-2008),求f'(1).用求导法则求.
f(x)=(x^1/2-1)(x^1/2-2)……(x^1/2-2008),求f'(1).用求导法则求.
f(x)=(√x-1)【余下的用g(x)表示】,则f'(x)=(√x-1)'g(x)+(√x-1)g'(x)=(√x-1)'g(x)=[1/(2√x)]g(x),代入计算下,结果是:-2007!/2
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