Question

If $y=e^x\cos x$, prove that $\frac{dy}{dx}=\sqrt{2}{e}^x·\cos \left(x+\frac{\pi }{4}\right)$

Answer 1

$\mathrm{We}\mathrm{have},y=e^x\cos x$
Differentiating with respect to x,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left(e^x\cos x\right) \\ =e^x\frac{d}{dx}\cos x+\cos x\frac{d}{dx}{e}^x \\ =e^x\left(-\sin x\right)+e^x\cos x \\ =e^x\left(\cos x-\sin x\right) \\ =\sqrt{2}{e}^x\left(\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}\right)\left[\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}\sqrt{2}\right] \\ =\sqrt{2}{e}^x\left(\cos \frac{\mathrm{\pi }}{4}\cos x-\sin \frac{\mathrm{\pi }}{4}\sin x\right) \\ =\sqrt{2}{e}^x\cos \left(x+\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{So},\frac{\mathit{d}y}{\mathit{d}x}=\sqrt{2}{e}^x\cos \left(x+\frac{\mathrm{\pi }}{4}\right) \end{gathered}$$