求第三题解答
1个回答
展开全部
(1)
y=f(x^2)+[f(x)]^2
dy/dx
=2xf'(x^2) + 2f(x) .f'(x)
(2)
y=x^2.f(cos(1/x))
dy/dx
=2x.f(cos(1/x)) + x^2 .f'(cos(1/x)) . ( -sin(1/x) ( -1/x^2)
=2x.f(cos(1/x)) + sin(1/x).f'(cos(1/x))
(3)
y=x^2.f(ax^2+b)
dy/dx
=2x.f(ax^2+b) + x^2. f'(ax^2+b) .(2ax)
=2x.f(ax^2+b) + 2ax^3. f'(ax^2+b)
(4)
y= f[g(lnx)]
dy/dx
=f'[g(lnx)] .g'(lnx). (1/x)
y=f(x^2)+[f(x)]^2
dy/dx
=2xf'(x^2) + 2f(x) .f'(x)
(2)
y=x^2.f(cos(1/x))
dy/dx
=2x.f(cos(1/x)) + x^2 .f'(cos(1/x)) . ( -sin(1/x) ( -1/x^2)
=2x.f(cos(1/x)) + sin(1/x).f'(cos(1/x))
(3)
y=x^2.f(ax^2+b)
dy/dx
=2x.f(ax^2+b) + x^2. f'(ax^2+b) .(2ax)
=2x.f(ax^2+b) + 2ax^3. f'(ax^2+b)
(4)
y= f[g(lnx)]
dy/dx
=f'[g(lnx)] .g'(lnx). (1/x)
追问
这个求出来是二阶导吗
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