(1)f ‘ (x^2)=1/x,x>0,求f(x)(2)g ' (x^2)=x^3,x>0,求g(x)

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(1)f ‘ (x^2)=1/x,x>0,求f(x)(2)g ' (x^2)=x^3,x>0,求g(x)
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(1)f ‘ (x^2)=1/x,x>0,求f(x)(2)g ' (x^2)=x^3,x>0,求g(x)
(1)f ‘ (x^2)=1/x,x>0,求f(x)
(2)g ' (x^2)=x^3,x>0,求g(x)

(1)f ‘ (x^2)=1/x,x>0,求f(x)(2)g ' (x^2)=x^3,x>0,求g(x)
f'(x^2)=1/x ) -> f'(x)=1/√(x) -> f(x)=∫f'(x)dx=∫1/√(x)dx=2√(x).
the same reasoning for the (2).

(1) f'(x^2)=1/x,x>0,求f(x)
令 u=x^2, x>0, 则 x=√u. f'(u)=1/√u, 得 f(u)=2√u+C, 即 f(x)=2√x+C.
(2) g’(x^2)=x^3,x>0,求g(x)
令 u=x^2, x>0, 则 x=√u. g'(u)=u^(3/2), 得 g(u)=(2/5)u^(5/2)+C, 即 g(x)...

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(1) f'(x^2)=1/x,x>0,求f(x)
令 u=x^2, x>0, 则 x=√u. f'(u)=1/√u, 得 f(u)=2√u+C, 即 f(x)=2√x+C.
(2) g’(x^2)=x^3,x>0,求g(x)
令 u=x^2, x>0, 则 x=√u. g'(u)=u^(3/2), 得 g(u)=(2/5)u^(5/2)+C, 即 g(x)=(2/5)x^(5/2)+C.

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