求导数的过程\``/``_```( x - √x`) `\/````````————`/```` ̄```( x + √x
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求导数的过程
\``/``_```( x - √x`)
`\/````````————
`/```` ̄```( x + √x`)
求y' ,
为什么书上的答案是这个的?:(√x)/(x+√x)^2
\``/``_```( x - √x`)
`\/````````————
`/```` ̄```( x + √x`)
求y' ,
为什么书上的答案是这个的?:(√x)/(x+√x)^2
![求导数的过程\``/``_```( x - √x`) `\/````````————`/```` ̄```( x + √x](/uploads/image/z/4061415-39-5.jpg?t=%E6%B1%82%E5%AF%BC%E6%95%B0%E7%9A%84%E8%BF%87%E7%A8%8B%5C%60%60%2F%60%60%EF%BC%BF%60%60%60%28+x+-+%E2%88%9Ax%60%29+%60%5C%2F%60%60%60%60%60%60%60%60%E2%80%94%E2%80%94%E2%80%94%E2%80%94%60%2F%60%60%60%60%EF%BF%A3%60%60%60%28+x+%2B+%E2%88%9Ax)
y = (x-√x)/(x+√x)..
y' = ( (x-√x)'(x+√x)-(x+√x)'(x-√x) ) / (x+√x)^2
= ( (1-x^(-1/2)/2)(x+√x) - (1+x^(-1/2)/2)(x-√x) ) / (x^2 + x + 2x√x)
= ( (1-x^(-1/2)/2)(1+x^(-1/2)) - (1+x^(-1/2)/2)(1-x^(-1/2)) ) / (1+x+2√x)
= ( (2-x^(-1/2))(1+x^(-1/2)) - (2+x^(-1/2))(1-x^(-1/2)) ) / 2(1+x+2√x)
= ( (2-x^(-1)+x^(-1/2)) - (2-x^(-1)-x^(-1/2)) ) / 2(1+x+2√x)
= (2x^(-1/2)) / (2(1+x+2√x)
= 1 / √x(1+x+2√x)
= 1 / (√x+2x+x√x)
y' = ( (x-√x)'(x+√x)-(x+√x)'(x-√x) ) / (x+√x)^2
= ( (1-x^(-1/2)/2)(x+√x) - (1+x^(-1/2)/2)(x-√x) ) / (x^2 + x + 2x√x)
= ( (1-x^(-1/2)/2)(1+x^(-1/2)) - (1+x^(-1/2)/2)(1-x^(-1/2)) ) / (1+x+2√x)
= ( (2-x^(-1/2))(1+x^(-1/2)) - (2+x^(-1/2))(1-x^(-1/2)) ) / 2(1+x+2√x)
= ( (2-x^(-1)+x^(-1/2)) - (2-x^(-1)-x^(-1/2)) ) / 2(1+x+2√x)
= (2x^(-1/2)) / (2(1+x+2√x)
= 1 / √x(1+x+2√x)
= 1 / (√x+2x+x√x)