求导结果 / Derivative Result
`d/dx ln(sqrt(x) + 1) = 1/(sqrt(x) + 1)*(1/(2*sqrt(x)) + 0)`
解题步骤 / Steps to Solution
我们知道, `d/dx ln(x) = 1/x`.
又由, `链式法则:dy/dx = dy/(du)(du)/dx`.
所以, `d/dx ln(sqrt(x) + 1) = 1/(sqrt(x) + 1)*d/dx (sqrt(x) + 1)`.
我们知道, `d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)`.
由此可得, `d/dx (sqrt(x) + 1) = d/dx sqrt(x) + d/dx 1`.
由, `d/dx sqrt(x) = 1/(2*sqrt(x))`.
由, `d/dx c = 0`.
可得, `d/dx 1 = 0`.
所以,根据定理:`d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)`,
`d/dx (sqrt(x) + 1) = 1/(2*sqrt(x)) + 0`
所以,根据法则, `d/dx ln(x) = 1/x`,
又因为, `链式法则:dy/dx = dy/(du)(du)/dx`,
`d/dx ln(sqrt(x) + 1) = 1/(sqrt(x) + 1)*(1/(2*sqrt(x)) + 0)`
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