Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (v) maths textbook solution

$27e^{6x}cos3x-36e^{6x}sin3x$

Hint:

You must know about derivative of cos3x & e6x

Given:

$e^{6x}cos3x$

Solution:

$Let\: \: y=e^{6x}cos3x$

Use multiplicative rule

$As\: \: UV=UV^{1}+U^{1}V$

Where U=e6x and V=cos3x

\begin{aligned} &\frac{d y}{d x}=e^{6 x} \frac{d}{d x} \cos 3 x+\frac{d}{d x} e^{6 x} \cdot \cos 3 x \\ &\frac{d y}{d x}=e^{6 x}-\sin 3 x \frac{d}{d x} 3 x+6 e^{6 x} \cos 3 x \quad\left(\frac{d \cos x}{d x}=-\sin x, \frac{d}{d x} e^{6 x}=e^{6 x} .6\right) \\ &\frac{d y}{d x}=-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x \\ &\frac{d^{2} y}{d x^{2}}=-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x \end{aligned}

$As\: \: UV=UV^{1}+U^{1}V$

Where U=e6x and V=sin3x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-3\left[e^{6 x} \cos 3 x(3)+e^{6 x} \cdot \sin 3 x .6\right]+6\left[e^{6 x} \cos 3 x \cdot 6+e^{6 x} \cdot(-\sin 3 x)\right] \\ &\left(\frac{d \sin 3 x}{d x}=3 \cos 3 x, \frac{d \cos 3 x}{d x}=3(-\sin 3 x), \frac{d e^{6 x}}{d x}=6 e^{6 x}\right) \\ &\frac{d^{2} y}{d x^{2}}=-3\left[3 e^{6 x} \cos 3 x+6 e^{6 x} \cdot \sin 3 x\right]+6\left[e^{6 x} \cos 3 x \cdot 6-3 e^{6 x} \cdot \sin 3 x\right] \\ &\frac{d^{2} y}{d x^{2}}=27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x \end{aligned}