有几个公式搞不懂是怎么推导出来的,我变换不出来:1.(sinx)'=cosx,2.(cosx)'= -sinx,3.(a^x)'=a^x lna,4.(e^x)'=e^x,5.[log(a)X]'=1/Xlna,6.(lnx)'=1/x,7.[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x).8.[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2.
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![有几个公式搞不懂是怎么推导出来的,我变换不出来:1.(sinx)'=cosx,2.(cosx)'= -sinx,3.(a^x)'=a^x lna,4.(e^x)'=e^x,5.[log(a)X]'=1/Xlna,6.(lnx)'=1/x,7.[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x).8.[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2.](/uploads/image/z/3168573-69-3.jpg?t=%E6%9C%89%E5%87%A0%E4%B8%AA%E5%85%AC%E5%BC%8F%E6%90%9E%E4%B8%8D%E6%87%82%E6%98%AF%E6%80%8E%E4%B9%88%E6%8E%A8%E5%AF%BC%E5%87%BA%E6%9D%A5%E7%9A%84%2C%E6%88%91%E5%8F%98%E6%8D%A2%E4%B8%8D%E5%87%BA%E6%9D%A5%EF%BC%9A1.%28sinx%29%27%3Dcosx%2C2.%28cosx%29%27%3D+-sinx%2C3.%28a%5Ex%29%27%3Da%5Ex+lna%2C4.%28e%5Ex%29%27%3De%5Ex%2C5.%5Blog%28a%29X%5D%27%3D1%2FXlna%2C6.%28lnx%29%27%3D1%2Fx%2C7.%5Bf%28x%29g%28x%29%5D%27%3Df%27%28x%29g%28x%29%2Bf%28x%29g%27%28x%29.8.%5Bf%28x%29%2Fg%28x%29%5D%27%3D%5Bf%27%28x%29g%28x%29-f%28x%29g%27%28x%29%5D%2F%5Bg%28x%29%5D%5E2.)
有几个公式搞不懂是怎么推导出来的,我变换不出来:1.(sinx)'=cosx,2.(cosx)'= -sinx,3.(a^x)'=a^x lna,4.(e^x)'=e^x,5.[log(a)X]'=1/Xlna,6.(lnx)'=1/x,7.[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x).8.[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2.
有几个公式搞不懂是怎么推导出来的,我变换不出来:
1.(sinx)'=cosx,
2.(cosx)'= -sinx,
3.(a^x)'=a^x lna,
4.(e^x)'=e^x,
5.[log(a)X]'=1/Xlna,
6.(lnx)'=1/x,
7.[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x).
8.[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2.
有几个公式搞不懂是怎么推导出来的,我变换不出来:1.(sinx)'=cosx,2.(cosx)'= -sinx,3.(a^x)'=a^x lna,4.(e^x)'=e^x,5.[log(a)X]'=1/Xlna,6.(lnx)'=1/x,7.[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x).8.[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2.
大哥,这么多,我帮你解决一个 思路都是一样的 就是采用导数的定义.
推导sinx,假设y趋向于0,则(sinx)'=[sin(x+y)-sinx]/(x+y-x)=[sin(x+y/2+y/2)-sin(x+y/2-y/2)]/y=
2[cos(x+y/2)][siny/2]/y,当y趋向于0时,cos(x+y/2)=cosx,siny/2等价于y/2,所以2[cos(x+y/2)][siny/2]/y=cosx
高中数学某公式的定义的推导,在高中范围内是不能推导出来的,你只要掌握用法就行了
用到极限的知识了,大概只需记住会用就好