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D dx (uv) = u dv dx v du dx g'(x) = {(x)( d dx e−x) (e−x)( d dx x)} ∴ g'(x) = {(x)( −e−x) (e−x)(1)} ∴ g'(x) = e−x − xe−x We can now write down the second derivative g''(x) = − e−x − (e−x − xe−x) ∴ g''(x) = − 2e−x xe−xЁ@ l b g N \ V x X ̉ߋ ̋ l R ~ ɗL l ͌Â Ȃ 폜 ܂ ł 鋁 l T ̕ 狁 l ɉ E ⍇ ł ܂Z q o g e c i t s u j e x e m s e c u r i t y a a x d c t e y r e b b o r k n a b g r a d p i s t o l u c c e e a i t a t r o p e r f r a u d r m s c v t o p s e c r e t j h i d a i n v e s t i g a t i o n e i i a d b w p b f f u t o a t b y e g m e f c x i o a v l i f l p z e c n e g i l l e t n i y r s j n o f f i c e r r u x

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