If-f-x-g-x-f-x-g-x-find-the-function-of-f-x-

Question Number 110456 by bobhans last updated on 29/Aug/20

I f {(f (x) . g (x))}^{'} = f {(x)}^{'} . g {(x)}^{'}

f i n d t h e f u n c t i o n o f f (x) .

Answered by bemath last updated on 29/Aug/20

\frac{d}{dx} (f (x) . g (x)) = f {(x)}^{'} . g (x) + f (x) . g {(x)}^{'}

then given condition

\Rightarrow f {(x)}^{'} . g (x) + f (x) . g^{'} (x) = f {(x)}^{'} . g {(x)}^{'}

\Rightarrow f {(x)}^{'} . g (x) - f {(x)}^{'} . g {(x)}^{'} = - f (x) . g {(x)}^{'}

\Rightarrow f {(x)}^{'} (g (x) - g {(x)}^{'}) = - f (x) . g {(x)}^{'}

\Rightarrow \frac{f {(x)}^{'}}{f (x)} = \frac{g {(x)}^{'}}{g {(x)}^{'} - g (x)}

\int \frac{f {(x)}^{'}}{f (x)} dx = \int \frac{g {(x)}^{'}}{g {(x)}^{'} - g (x)} dx

\Rightarrow \ln (f (x)) = \int \frac{g {(x)}^{'}}{g {(x)}^{'} - g (x)} dx + c

\Rightarrow f (x) = C . \exp (\int \frac{g {(x)}^{'}}{g {(x)}^{'} - g (x)} dx)

∴ f (x) = C . e^{\int (\frac{g {(x)}^{'}}{g {(x)}^{'} - g (x)} dx)}

Commented by bobhans last updated on 29/Aug/20

great \dots .

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