Hello guys can someone help me with that?
I tried all the ways but it always leads me far away from the wanted result.
Given $g(x)= e^{x}-x-1$ and $f(x)=x-1+(x+2)e^{-x}$, prove that $f'(x) = e^{-x} g(x)$.
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well after all I did, it ended up with f'(x) =e^(-x)+1- e^(-x) (x+2) – chxmxii Apr 22 '19 at 20:04
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sorry but how to do this e^x? – chxmxii Apr 22 '19 at 20:09
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Your answer is correct. We can factorise it as follows; $$f'(x)=1+e^{-x}(-x-1)=e^{-x}(e^x-x-1)=e^{-x}g(x)$$
Peter Foreman
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$e^{-x}\times e^x=e^{-x+x}=e^0=1$. So we can factorise $1$ as $1=e^{-x}\times e^x$. – Peter Foreman Apr 22 '19 at 20:12