Question
If $\text{y}=\text{e}^{\text{x}}\cos\text{x},$ Prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}.\cos\Big(\text{x}+\frac{\pi}{4}\Big)$

Answer

Given, $\text{y}=\text{e}^{\text{x}}\cos\text{x}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\cos\text{x}\big)$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}$ [Using product rule]
$=\text{e}^\text{e}(-\sin\text{x})+\text{e}^\text{x}\cos\text{x}$
$=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
$=\sqrt{2}\text{e}^\text{x}\Big(\frac{\cos\text{x}}{\sqrt{2}}-\frac{\sin\text{x}}{\sqrt{2}}\Big)$ $\big[$Multiplying and dividing by $\sqrt{2}\big]$
$=\sqrt{2}\text{e}^\text{x}\Big(\cos\frac{\pi}{4}\cos\text{x}-\sin\frac{\pi}{4}\sin\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}\cos\Big(\text{x}+\frac{\pi}{4}\Big)$

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