不知道这个不等式的导数怎么得来的!设f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)f'(x)=ln[x+√(1+x²)]+ x*[1+x/√(1+x²)]-x/√(1+x²)=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)=ln[x+√(1+x²)]谁
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![不知道这个不等式的导数怎么得来的!设f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)f'(x)=ln[x+√(1+x²)]+ x*[1+x/√(1+x²)]-x/√(1+x²)=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)=ln[x+√(1+x²)]谁](/uploads/image/z/13369967-71-7.jpg?t=%E4%B8%8D%E7%9F%A5%E9%81%93%E8%BF%99%E4%B8%AA%E4%B8%8D%E7%AD%89%E5%BC%8F%E7%9A%84%E5%AF%BC%E6%95%B0%E6%80%8E%E4%B9%88%E5%BE%97%E6%9D%A5%E7%9A%84%21%E8%AE%BEf%28x%29%3Dxln%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%2B1-%E2%88%9A%281%2Bx%26%23178%3B%29%2C%28+x%3E0%29f%27%28x%29%3Dln%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%2B+x%2A%5B1%2Bx%2F%E2%88%9A%281%2Bx%26%23178%3B%29%5D-x%2F%E2%88%9A%281%2Bx%26%23178%3B%29%3Dln%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%2Bx%2F%E2%88%9A%281%2Bx%26%23178%3B%29-x%2F%E2%88%9A%281%2Bx%26%23178%3B%29%3Dln%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%E8%B0%81)
不知道这个不等式的导数怎么得来的!设f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)f'(x)=ln[x+√(1+x²)]+ x*[1+x/√(1+x²)]-x/√(1+x²)=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)=ln[x+√(1+x²)]谁
不知道这个不等式的导数怎么得来的!
设f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)
f'(x)=ln[x+√(1+x²)]+ x*[1+x/√(1+x²)]-x/√(1+x²)
=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)
=ln[x+√(1+x²)]
谁可以再写详细点?我怎么算都没办法算出最后是ln[x+√(1+x²)]
不知道这个不等式的导数怎么得来的!设f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)f'(x)=ln[x+√(1+x²)]+ x*[1+x/√(1+x²)]-x/√(1+x²)=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)=ln[x+√(1+x²)]谁
f(x)=xln[x+√(1+x²)]+1-√(1+x²),( x>0)
f'(x)=ln[x+√(1+x²)]
+ x*[1+x/√(1+x²)] *[1/(x+√(1+x²)] 这个地方是关键吧
-x/√(1+x²)
=ln[x+√(1+x²)]
+x/√(1+x²)
-x/√(1+x²)
=ln[x+√(1+x²)]
g(x)=xln[x+√(1+x²)]
g'(x)=ln[x+√(1+x²)]+ x *1/[x+√(1+x²)] *(1+x/√(1+x²))
=ln[x+√(1+x²)]+ x*(1/[x+√(1+x²)] *([x+√(1+x²)]/ √(1+x²))
=ln[x+√(...
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g(x)=xln[x+√(1+x²)]
g'(x)=ln[x+√(1+x²)]+ x *1/[x+√(1+x²)] *(1+x/√(1+x²))
=ln[x+√(1+x²)]+ x*(1/[x+√(1+x²)] *([x+√(1+x²)]/ √(1+x²))
=ln[x+√(1+x²)]+ x/√(1+x²)
f'(x)=g'(x)-(√(1+x²))'=ln[x+√(1+x²)]+ x/√(1+x²) -x/√(1+x²)=ln[x+√(1+x²)]
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答案是没错的,过程中有一个地方错了啊。就是第一步求导的第二个式子。
改正后应该是:
f'(x)=ln[x+√(1+x²)]+ x*1/[x+√(1+x²)]*[1+x/√(1+x²)]-x/√(1+x²)
=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)
=ln[x+√(1+x...
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答案是没错的,过程中有一个地方错了啊。就是第一步求导的第二个式子。
改正后应该是:
f'(x)=ln[x+√(1+x²)]+ x*1/[x+√(1+x²)]*[1+x/√(1+x²)]-x/√(1+x²)
=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)
=ln[x+√(1+x²)]
这就对了。(它漏掉了东西)
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