#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (iv) maths

$-24e^{x}sin\: 5x+10e^{x}cos\: 5x$

Hint:

You must know about derivative of sin x & ex

Given:

$e^{x}sin\: 5x$

Solution:

$Let\: \: y=e^{x}sin\: 5x$

Use multiplicative rule

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

Where U=ex & V=sin 5x

\begin{aligned} &\frac{d y}{d x}=e^{x} \frac{d}{d x} \sin 5 x+\frac{d}{d x} e^{x} \cdot \sin 5 x \\ &\frac{d y}{d x}=e^{x} \cdot \cos 5 x \frac{d}{d x} 5 x+e^{x} \cdot \sin 5 x \quad\left(\frac{d}{d x} e^{x}=e^{x}\right) \quad\left(\frac{d \sin 5 x}{d x}=\cos 5 x\right) \\ &\frac{d y}{d x}=e^{x} \cdot \cos 5 x \cdot 5+e^{x} \cdot \sin 5 x \quad \quad\left(\frac{d}{d x} 5 x=5\right) \\ &\frac{d y}{d x}=5 e^{x} \cos 5 x+e^{x} \sin 5 x \end{aligned}

\begin{aligned} &\frac{d}{d x}(\frac{d y}{d x})=\frac{d}{d x}(5 e^{x} \cos 5 x+e^{x} \sin 5 x) \end{aligned}

Again use multiplication rule

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

Where U=ex & V=sin 5x

U=ex & V=cos 5x

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \frac{d}{d x} \cos 5 x+\frac{d}{d x} e^{x} \cdot \cos 5 x\right]+\left[e^{x} \frac{d}{d x} \sin 5 x+\frac{d}{d x} e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \cdot(-\sin 5 x) \frac{d}{d x} 5 x+e^{x} \cdot \cos 5 x\right]+\left[e^{x} \cos 5 x \frac{d}{d x} 5 x+e^{x} \cdot \sin 5 x\right] \\ &\left(\frac{d \sin x}{d x}=\cos x, \frac{d \cos x}{d x}=-\sin x, \frac{d 5 x}{d x}=5\right) \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \cdot(-\sin 5 x) 5+e^{x} \cdot \cos 5 x\right]+\left[e^{x} \cos 5 x .5+e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=\left[25 e^{x} \sin 5 x+5 e^{x} \cdot \cos 5 x\right]+\left[5 e^{x} \cos 5 x +e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=-24 e^{x} \sin 5 x+10 e^{x} \cdot \cos 5 x \end{aligned}