What is expand f(x)x(2(y)''-x(y)'=e^{-x}) ?

The solution to expand f(x)x(2(y)''-x(y)'=e^{-x}) is 2x^2f(x)(y)''-x^3f(x)(y)'=e^{-x}

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expand f(x)x(2(y)''-x(y)'=e^{-x})

expand

f (x) xx (2 (y)^{''} - x (y)^{'} = e^{- x})

2 x^{2} f (x) (y)'' - x^{3} f (x) (y)' = e^{- x}

Solution steps

f (x) xx (2 (y)^{''} - x (y)^{'} = e^{- x})

= xx f (x) (2 (y)^{^{''}} - = e^{- x} x (y)^{^{'}})

Apply the distributive law:

a (b - c) = ab - a c

a = f (x) xx, b = 2 (y)^{''}, c = x (y)^{'} = e^{- x}

f (x) xx \cdot 2 (y)^{^{''}} - f (x) xxx (y)^{^{'}} = e^{- x}

2 xx f (x) (y)^{^{''}} - xxx f (x) (y)^{^{'}} = e^{- x}

Expand

2 xx f (x) (y)^{''} - xxx f (x) (y)^{'} : 2 x^{2} f (x) (y)^{''} - x^{3} f (x) (y)^{'}

2 x^{2} f (x) (y)^{^{''}} - x^{3} f (x) (y)^{^{'}} = e^{- x}

e^{x} (x + e^{x})

\frac{x ( 2 - x ) ^{4}}{4}

(x + sin (x) - 1)^{2}

(2 x) (3 x + 1 + 1)

(- \frac{1}{2} e^{t} + \frac{3}{2} e^{- t}) \cdot (4 e^{t} + e^{- t})

What is expand f(x)x(2(y)''-x(y)'=e^{-x}) ?
The solution to expand f(x)x(2(y)''-x(y)'=e^{-x}) is 2x^2f(x)(y)''-x^3f(x)(y)'=e^{-x}