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Question
If \[y = e^x \cos x\] ,prove that \[\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)\] ?
Solution
\[\text{ We have, y } = e^x \cos x\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( e^x \cos x \right)\]
\[ = e^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} e^x \]
\[ = e^x \left( - \sin x \right) + e^x \cos x\]
\[ = e^x \left( \cos x - \sin x \right)\]
\[ = \sqrt{2} e^x \left( \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} \right) \left[ \text{Multiplying and dividing by } \sqrt{2} \right]\]
\[ = \sqrt{2} e^x \left( \cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x \right)\]
\[ = \sqrt{2} e^x \cos\left( x + \frac{\pi}{4} \right)\]
\[So, \frac{d y}{d x} = \sqrt{2} e^x \cos\left( x + \frac{\pi}{4} \right)\]
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