导数的乘法法则推倒uv)'=lim(h→0)[u(x+h)v(x+h)-uv]/h=lim(h→0)[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h=lim(h→0)[u(x+h)]×[v(x+h)-v(x)]/h+lim(h→0)[v(x)]×[u(x+h)-u(x)]/h=u(x)v'(x)+u'(x)v(x)=u'v+uv'请问这个第一步lim(h→0)[u(x+h)v(x+h)-
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 14:17:45
![导数的乘法法则推倒uv)'=lim(h→0)[u(x+h)v(x+h)-uv]/h=lim(h→0)[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h=lim(h→0)[u(x+h)]×[v(x+h)-v(x)]/h+lim(h→0)[v(x)]×[u(x+h)-u(x)]/h=u(x)v'(x)+u'(x)v(x)=u'v+uv'请问这个第一步lim(h→0)[u(x+h)v(x+h)-](/uploads/image/z/1069867-19-7.jpg?t=%E5%AF%BC%E6%95%B0%E7%9A%84%E4%B9%98%E6%B3%95%E6%B3%95%E5%88%99%E6%8E%A8%E5%80%92uv%29%27%3Dlim%28h%E2%86%920%29%5Bu%28x%2Bh%29v%28x%2Bh%29-uv%5D%2Fh%3Dlim%28h%E2%86%920%29%5Bu%28x%2Bh%29v%28x%2Bh%29%2Bu%28x%2Bh%29v-u%28x%2Bh%29v-uv%5D%2Fh%3Dlim%28h%E2%86%920%29%5Bu%28x%2Bh%29%5D%C3%97%5Bv%28x%2Bh%29-v%28x%29%5D%2Fh%2Blim%28h%E2%86%920%29%5Bv%28x%29%5D%C3%97%5Bu%28x%2Bh%29-u%28x%29%5D%2Fh%3Du%28x%29v%27%28x%29%2Bu%27%28x%29v%28x%29%3Du%27v%2Buv%27%E8%AF%B7%E9%97%AE%E8%BF%99%E4%B8%AA%E7%AC%AC%E4%B8%80%E6%AD%A5lim%28h%E2%86%920%29%5Bu%28x%2Bh%29v%28x%2Bh%29-)
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来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 14:17:45