Question

If y = e^x cos x, prove that \[\frac{d^2 y}{d x^2} = 2 e^x \cos \left( x + \frac{\pi}{2} \right)\] ?

Answer 1

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

Hence proved.