Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 67 maths textbook solution

Hint: you must know the rules of derivative of exponential and trigonometric functions.

Given: $y=e^{x} \cos x$

Prove: $\frac{d y}{d x}=\sqrt{2} e^{x} \cdot \cos \left(x+\frac{\pi}{4}\right)$

Solution:

Let  $y=e^{x} \cos x$

Differentiate with respect to x,use product rule

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(e^{x} \cos x\right) \\\\ &\frac{d y}{d x}=e^{x} \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x} e^{x} \end{aligned}

\begin{aligned} &\frac{d y}{d x}=e^{x}(-\sin x)+e^{x}(\cos x) \\\\ &\frac{d y}{d x}=e^{x}(\cos x-\sin x) \end{aligned}

Multiply and divide by$\sqrt{2}$

\begin{aligned} &\frac{d y}{d x}=\sqrt{2} e^{x}\left(\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}\right) \\\\ &\frac{d y}{d x}=\sqrt{2} e^{x}\left(\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x\right) \end{aligned}

$\frac{d y}{d x}=\sqrt{2} e^{x} \cos \left(x+\frac{\pi}{4}\right)$               [ ∴using property  $\operatorname{cos \; a\; cos} b-\operatorname{sin\: a\; sin} b=\cos (a+b)$]

∴ Proved

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