求y=(x-1)(x-2)(x-3)...(x-10)(x大于10)的导数.f(x)=(x-1)(x-2)……(x-10),ln[f(x)]=ln[(x-1)(x-2)……(x-10)]ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10){ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(x-10)]'f'(x)/f(x)=1/(x-1)+1/(x-2)+……+1/(x-10)f'(
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![求y=(x-1)(x-2)(x-3)...(x-10)(x大于10)的导数.f(x)=(x-1)(x-2)……(x-10),ln[f(x)]=ln[(x-1)(x-2)……(x-10)]ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10){ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(x-10)]'f'(x)/f(x)=1/(x-1)+1/(x-2)+……+1/(x-10)f'(](/uploads/image/z/615479-23-9.jpg?t=%E6%B1%82y%3D%28x-1%29%28x-2%29%28x-3%29...%EF%BC%88x-10%29%EF%BC%88x%E5%A4%A7%E4%BA%8E10%EF%BC%89%E7%9A%84%E5%AF%BC%E6%95%B0.f%28x%29%3D%28x-1%29%28x-2%29%E2%80%A6%E2%80%A6%28x-10%29%2Cln%5Bf%28x%29%5D%3Dln%5B%28x-1%29%28x-2%29%E2%80%A6%E2%80%A6%28x-10%29%5Dln%5Bf%28x%29%5D%3Dln%28x-1%29%2Bln%28x-2%29%2B%E2%80%A6%E2%80%A6%2Bln%28x-10%29%7Bln%5Bf%28x%29%5D%7D%27%3D%5Bln%28x-1%29%2Bln%28x-2%29%2B%E2%80%A6%E2%80%A6%2Bln%28x-10%29%5D%27f%27%28x%29%2Ff%28x%29%3D1%2F%28x-1%29%2B1%2F%28x-2%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28x-10%29f%27%28)
求y=(x-1)(x-2)(x-3)...(x-10)(x大于10)的导数.f(x)=(x-1)(x-2)……(x-10),ln[f(x)]=ln[(x-1)(x-2)……(x-10)]ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10){ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(x-10)]'f'(x)/f(x)=1/(x-1)+1/(x-2)+……+1/(x-10)f'(
求y=(x-1)(x-2)(x-3)...(x-10)(x大于10)的导数.
f(x)=(x-1)(x-2)……(x-10),
ln[f(x)]=ln[(x-1)(x-2)……(x-10)]
ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10)
{ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(x-10)]'
f'(x)/f(x)=1/(x-1)+1/(x-2)+……+1/(x-10)
f'(x)=[1/(x-1)+1/(x-2)+……+1/(x-10)]f(x)
我想问下[ln(x-1)+ln(x-2)+……+ln(x-10)]'为何是1/(x-1)+1/(x-2)+……+1/(x-10)
不是要进行二次求导么?ln(x-2)'应该是1/(2-x)啊
求y=(x-1)(x-2)(x-3)...(x-10)(x大于10)的导数.f(x)=(x-1)(x-2)……(x-10),ln[f(x)]=ln[(x-1)(x-2)……(x-10)]ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10){ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(x-10)]'f'(x)/f(x)=1/(x-1)+1/(x-2)+……+1/(x-10)f'(
[ln(x-2)]'=[1/(x-2)](x-2)'=1/(x-2)
你的那个负号是哪来的?你做错了.
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