Question

设f（x）可导，求下列函数的导数。大一高数。 麻烦写一下详细过程？

Answer 1

(1)
y=[f(x)]^2
y'
=2f(x) .(f(x))'
=2f(x) .f'(x)
(2)
y=e^[f(x)]
y'
=e^[f(x)] . (f(x))'
=e^[f(x)] . f'(x)
(3)
y=f(x^2)
y'
=f'(x^2) .(x^2)'
=f'(x^2) .(2x)
=2x.f'(x^2)
(4)
y=ln{1+[f(e^x)]^2}
y'
=【1/{1+[f(e^x)]^2}】.{1+[f(e^x)]^2}'
=【1/{1+[f(e^x)]^2}】. {2f(e^x)]}. (f(e^x))'
=【1/{1+[f(e^x)]^2}】. [2f(e^x)]. f'(e^x) .(e^x)'
=【1/{1+[f(e^x)]^2}】. [2f(e^x)]. f'(e^x) .(e^x)
=2e^x. f(e^x).f'(e^x)/{1+[f(e^x)]^2}

Answer 2

函数计算还不in年代开始看你很近百户那你能否科技

Answer 3

你应该去看一下复合函数的倒数。 比较简单，但不好理解。 1. y'=f'(2x+1)*2 2. y'=2[xf(x^2)]*[xf(x^2)]'=2[xf(x^2)]*[f(x^2)+x*f'(x^2)*2x] 3. 4.

Answer 4

用链式法则 u=f(x)u'=f'(x)y=u³所以 dy/dx==3u²*u'=3f²(x)*f'(x)