#### Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (ii)

$\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}}$

Hint:

You must know about derivative of sin x & log x

Given:

$sin(log\: x)$

Solution:

$Let\: \: y=sin(log\: x)$

\begin{aligned} &\left.\frac{d y}{d x}=\cos (\log x) \times \frac{d}{d x} \log x \quad \text { ( } \frac{d \sin x}{d x}=\cos x \right) \\ &\left.\frac{d y}{d x}=\cos (\log x) \times \frac{1}{x} \quad \text { ( } \frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d y}{d x}=\frac{\cos (\log x)}{x} \end{aligned}

Use quotient rule

$As\: \: \frac{u}{v}=\frac{u'v-v'u}{v^{2}}$

$W\! here\: \: v=x\: \, and \: \: u=cos(log\: x)$

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\left.x \frac{d}{d x}(\cos (\log x))-\cos (\log x)\right) \frac{d}{d x}(x)}{x^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-x \sin (\log x) \cdot \frac{d(\log x)}{d x}-1 \cdot \cos (\log x)}{x^{2}}\left(\frac{d}{d x} x=1\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sin (\log x) \cdot \frac{1}{x} x-\cos (\log x)}{x^{2}} \quad\left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sin (\log x)-\cos (\log x)}{x^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}} \end{aligned}