作业帮y=ln(x+√1+X^2)的导数 y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²)) * [x+√(1+x²)]'=1/(x+√(1+x²)) * [1+2x/2√(1+x²)][x+√(1+x²)]'→[1+2x/2√(1+x²)] 这一步到后一步是怎么求出来的
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/05 11:32:57
![y=ln(x+√1+X^2)的导数 y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²)) * [x+√(1+x²)]'=1/(x+√(1+x²)) * [1+2x/2√(1+x²)][x+√(1+x²)]'→[1+2x/2√(1+x²)] 这一步到后一步是怎么求出来的](/uploads/image/z/1042714-10-4.jpg?t=y%3Dln%28x%2B%E2%88%9A1%2BX%5E2%29%E7%9A%84%E5%AF%BC%E6%95%B0+y%27%3D%5Bln%28x%2B%E2%88%9A%281%2Bx%26%23178%3B%29%29%5D%27%3D1%2F%28x%2B%E2%88%9A%281%2Bx%26%23178%3B%29%29+%2A+%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%27%3D1%2F%28x%2B%E2%88%9A%281%2Bx%26%23178%3B%29%29+%2A+%5B1%2B2x%2F2%E2%88%9A%281%2Bx%26%23178%3B%29%5D%5Bx%2B%E2%88%9A%281%2Bx%26%23178%3B%29%5D%27%E2%86%92%5B1%2B2x%2F2%E2%88%9A%281%2Bx%26%23178%3B%29%5D+%E8%BF%99%E4%B8%80%E6%AD%A5%E5%88%B0%E5%90%8E%E4%B8%80%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E6%B1%82%E5%87%BA%E6%9D%A5%E7%9A%84)
y=ln(x+√1+X^2)的导数 y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²)) * [x+√(1+x²)]'=1/(x+√(1+x²)) * [1+2x/2√(1+x²)][x+√(1+x²)]'→[1+2x/2√(1+x²)] 这一步到后一步是怎么求出来的
y=ln(x+√1+X^2)的导数
y'=[ln(x+√(1+x²))]'
=1/(x+√(1+x²)) * [x+√(1+x²)]'
=1/(x+√(1+x²)) * [1+2x/2√(1+x²)]
[x+√(1+x²)]'→[1+2x/2√(1+x²)] 这一步到后一步是怎么求出来的
y=ln(x+√1+X^2)的导数 y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²)) * [x+√(1+x²)]'=1/(x+√(1+x²)) * [1+2x/2√(1+x²)][x+√(1+x²)]'→[1+2x/2√(1+x²)] 这一步到后一步是怎么求出来的
[x+√(1+x²)]'=x'+[√(1+x²)]'=1+[√(1+x²)]'
关键是后面的[√(1+x²)]'如何计算,用链式法则
令y=√(1+x²), u=1+x², 则
y=√u
∴y'=dy/dx
=(dy/du)*(du/dx)
=[d(√u)/du]*[d(1+x²)/dx]
=[1/(2√u)]*(2x)
=2x/2√u
=2x/2√(1+x²)
=x/√(1+x²)
∴[x+√(1+x²)]'=x'+[√(1+x²)]'=1+[√(1+x²)]'=1+x/√(1+x²)
[x+√(1+x²)]'
=1+(1/2)(1+x^2)^(-1/2)*2x
=1+(2x/2)(1+x^2)^(-1/2)
=1+x/√(1+x²)
根号下1+X2是复合函数, 可以化成指数函数,求导后指数是1-1/2=-1/2,所以根式在分母,余下的还要对1+X2求导