设f(x)与g(x)可导,f^2(x)+g^2 (x)≠0,求证函数y=根号下f^2(x)+g^2 (x)可导
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设f(x)与g(x)可导,f^2(x)+g^2 (x)≠0,求证函数y=根号下f^2(x)+g^2 (x)可导
设f(x)与g(x)可导,f^2(x)+g^2 (x)≠0,求证函数y=根号下f^2(x)+g^2 (x)可导
设f(x)与g(x)可导,f^2(x)+g^2 (x)≠0,求证函数y=根号下f^2(x)+g^2 (x)可导
△x=h
△y/△x=[√(f^2(x+h)+g^2 (x+h))-√(f^2(x)+g^2 (x))]/h
分子有理化
△y/△x=[(f^2(x+h)+g^2 (x+h))-(f^2(x)+g^2 (x))]
/[h(√(f^2(x+h)+g^2 (x+h))+√(f^2(x)+g^2 (x)))] (分子有理化)
=[(f^2(x+h) -f^2(x))+(g^2 (x+h)- g^2 (x))]
/[h(√(f^2(x+h)+g^2 (x+h))+√(f^2(x)+g^2 (x)))](集项)
=[(f(x+h) -f (x))( f(x+h) +f (x))/ h +(g (x+h)- g (x)) (g (x+h)+ g (x))/h]*
1/[ (√(f^2(x+h)+g^2 (x+h))+√(f^2(x)+g^2 (x))](平方差分解因式)
f(x)可导,f’(x)存在,
且f(x)连续,故当h→0时,f(x+h)→f (x)
于是当h→0时,
(f(x+h) -f (x))( f(x+h) +f (x))/ h→2f’(x)f(x)
同理当h→0时,
(g (x+h)-g (x))(g (x+h)+g (x))/h→2g’(x)g(x)
同理当h→0时,
1/[ (√(f^2(x+h)+g^2 (x+h))+√(f^2(x)+g^2 (x))] →1/[2√(f^2(x)+g^2 (x))]
(有意义,因为f^2(x)+g^2 (x)≠0)
从而
(△x→0)lim(△y/△x)=[ f’(x)f(x) +g’(x)g(x)]/ √(f^2(x)+g^2 (x))