求y=(1+ax)^x的导数(a为常数)有两种方法:\x051.y ̇=e^(xln(1+ax))=e^(xln(1+ax))[ln(1+ax)+ax/(1+ax)]= (1+ax)^x[ln(1+ax)+ax/(1+ax)]\x052.y ̇=a(1+ax)^xln(1+ax) (根据y=b^x的导数为y ̇=b^xlnb)请问上面两种方
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![求y=(1+ax)^x的导数(a为常数)有两种方法:\x051.y ̇=e^(xln(1+ax))=e^(xln(1+ax))[ln(1+ax)+ax/(1+ax)]= (1+ax)^x[ln(1+ax)+ax/(1+ax)]\x052.y ̇=a(1+ax)^xln(1+ax) (根据y=b^x的导数为y ̇=b^xlnb)请问上面两种方](/uploads/image/z/7004567-47-7.jpg?t=%E6%B1%82y%3D%281%2Bax%29%5Ex%E7%9A%84%E5%AF%BC%E6%95%B0%28a%E4%B8%BA%E5%B8%B8%E6%95%B0%29%E6%9C%89%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%B3%95%EF%BC%9A%5Cx051.y+%26%23775%3B%3De%5E%28xln%EF%BC%881%2Bax%EF%BC%89%29%3De%5E%28xln%EF%BC%881%2Bax%EF%BC%89%29%5Bln%281%2Bax%29%2Bax%2F%281%2Bax%29%5D%3D+%281%2Bax%29%5Ex%5Bln%281%2Bax%29%2Bax%2F%281%2Bax%29%5D%5Cx052.y+%26%23775%3B%3Da%281%2Bax%29%5Exln%281%2Bax%29+%EF%BC%88%E6%A0%B9%E6%8D%AEy%3Db%5Ex%E7%9A%84%E5%AF%BC%E6%95%B0%E4%B8%BAy+%26%23775%3B%3Db%5Exlnb%EF%BC%89%E8%AF%B7%E9%97%AE%E4%B8%8A%E9%9D%A2%E4%B8%A4%E7%A7%8D%E6%96%B9)
求y=(1+ax)^x的导数(a为常数)有两种方法:\x051.y ̇=e^(xln(1+ax))=e^(xln(1+ax))[ln(1+ax)+ax/(1+ax)]= (1+ax)^x[ln(1+ax)+ax/(1+ax)]\x052.y ̇=a(1+ax)^xln(1+ax) (根据y=b^x的导数为y ̇=b^xlnb)请问上面两种方
求y=(1+ax)^x的导数(a为常数)
有两种方法:
\x051.y ̇=e^(xln(1+ax))=e^(xln(1+ax))[ln(1+ax)+ax/(1+ax)]= (1+ax)^x[ln(1+ax)+ax/(1+ax)]
\x052.y ̇=a(1+ax)^xln(1+ax) (根据y=b^x的导数为y ̇=b^xlnb)
请问上面两种方法哪种是正确的,哪种是错误的,错误的为什么错了,答案是给的第一种,可我觉得第二种也对,可为什么结果不一样呢,
求y=(1+ax)^x的导数(a为常数)有两种方法:\x051.y ̇=e^(xln(1+ax))=e^(xln(1+ax))[ln(1+ax)+ax/(1+ax)]= (1+ax)^x[ln(1+ax)+ax/(1+ax)]\x052.y ̇=a(1+ax)^xln(1+ax) (根据y=b^x的导数为y ̇=b^xlnb)请问上面两种方
底数和指数都含有未知数,和导数基本公式不符,第二个方法不能直接用.第一种方法避免了这个问题