Question

If y = e^−x cos x, show that \[\frac{d^2 y}{d x^2} = 2 e^{- x} \sin x\] ?

Answer 1

Here,

\[y = e^{- x} \cos x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = - e^{- x} \sin x - e^{- x} \cos x\]
\[ = - \left( e^{- x} \sin x + e^{- x} \cos x \right)\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = - \left( e^{- x} \cos x - e^{- x} \sin x - e^{- x} \sin x - e^{- x} \cos x \right)\]
\[ = 2 e^{- x} \sin x\]

Hence proved.