If $y = e^{x} sin x$ and $z = e^{x} cos x$ then

y^{'} = z

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y^{''} = 2 z

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z^{''} = - 2 y

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z^{'} = - y

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Solution

The correct options are
A $y^{''} = 2 z$
C $z^{''} = - 2 y$
$y = e^{x} s i n x y^{'} = e^{x} s i n x + e^{x} c o s x y^{''} = e^{x} s i n x + e^{x} c o s x + e^{x} c o s x - e^{x} s i n x = 2 e^{x} c o s x = 2 z z = e^{x} c o s x z^{'} = e^{x} c o s x - e^{x} s i n x z^{''} = e^{x} c o s x - e^{x} s i n x - e^{x} s i n x - e^{x} c o s x = - 2 e^{x} s i n x = - 2 y$

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