Q95 / 200 · Manipal (MET) 2014 · Mathematics
If $y=|\sin x|^{|x|}$, then the value of $\frac{d y}{d x}$ at $x=\frac{-\pi}{6}$ is
- Correct answerA$\frac{2^{-\frac{\pi}{6}}}{}[6 \log 2-\sqrt{3} \pi]$
- B$2^{\frac{\pi}{6}}[6 \log 2+\sqrt{3} \pi]$
- C$\frac{2^{-\frac{\pi}{6}}}{6}[6 \log 2+\sqrt{3} \pi]$
- DNone of the above
Solution
Given $y=|\sin x|^{|x|}$
In the neighbourhood of $-\frac{\pi}{6},|x|$ and $|\sin x|$ both are negative
i.e., $y=(-\sin x)^{-x}$
Taking log on both sides, we get $\log y=-x \log (-\sin x)$ $\begin{aligned} & \begin{aligned} \Rightarrow \frac{1}{y} \frac{d y}{d x}= & (-x) \frac{1}{-\sin x}(-\cos x)+\log (-\sin x)(-1) \\ & =-x \cdot \cot x-\log (-\sin x) \\ & =-[x \cot x+\log (-\sin x)]\end{aligned} \\ & \begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =-y[x \cot x+\log (-\sin x)]\end{aligned} \\ & \begin{aligned} \therefore\left(\frac{d y}{d x}\right)_{x=-\frac{\pi}{6}}=\frac{2^{-\frac{\pi}{6}}}{}[6 \log 2-\sqrt{3} \pi]\end{aligned}\end{aligned}$
Taking log on both sides, we get $\log y=-x \log (-\sin x)$ $\begin{aligned} & \begin{aligned} \Rightarrow \frac{1}{y} \frac{d y}{d x}= & (-x) \frac{1}{-\sin x}(-\cos x)+\log (-\sin x)(-1) \\ & =-x \cdot \cot x-\log (-\sin x) \\ & =-[x \cot x+\log (-\sin x)]\end{aligned} \\ & \begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =-y[x \cot x+\log (-\sin x)]\end{aligned} \\ & \begin{aligned} \therefore\left(\frac{d y}{d x}\right)_{x=-\frac{\pi}{6}}=\frac{2^{-\frac{\pi}{6}}}{}[6 \log 2-\sqrt{3} \pi]\end{aligned}\end{aligned}$
Hence, the correct answer is option: 'A'.