g’(y)=1/f’(x)那么g‘’(y)等于？写详细一点😩
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如下:g'(y) = 1/f'(x)
d/dx [g'(y)] =d/dx[ 1/f'(x)]
g''(y) . y' = -f''(x)/[f'(x)]^2
g''(y) = -f''(x)/{ [f'(x)]^2 .y' }。
d/dx [g'(y)] =d/dx[ 1/f'(x)]
g''(y) . y' = -f''(x)/[f'(x)]^2
g''(y) = -f''(x)/{ [f'(x)]^2 .y' }。
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g'(y) = 1/f'(x)
d/dx [g'(y)] =d/dx[ 1/f'(x)]
g''(y) . y' = -f''(x)/[f'(x)]^2
g''(y) = -f''(x)/{ [f'(x)]^2 .y' }
d/dx [g'(y)] =d/dx[ 1/f'(x)]
g''(y) . y' = -f''(x)/[f'(x)]^2
g''(y) = -f''(x)/{ [f'(x)]^2 .y' }
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g'(y)=1/f'(x)
g''(y)=[1/f'(x)]'=-[f'(x)]'/[f'(x)]^2=-f''(x)/[f'(x)]^2
g''(y)=[1/f'(x)]'=-[f'(x)]'/[f'(x)]^2=-f''(x)/[f'(x)]^2
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