If y = | x | ^{ | x | } then what is the value of dy/dx at x=-π/6?

Correct Answer is : 2−π6(6ln2−3π)6{2^{-π /6}(6ln2-{3}π )}{6}62−6π(6ln2−3π)

Correct Answer is : 2−π6(6ln2−3π)6{2^{-π /6}(6ln2-{3}π )}{6}62−6π(6ln2−3π)

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UPSC NDA II 2018 Mathematics Solved Paper

Question : 76 of 120

Marks: +1, -0

y = | \sin x | ^{ | x | }

\frac{dy}{dx}

x=-\frac{\pi}{6}

Solution:

Concept:

For

x = -\frac{\pi}{6} < 0

, the absolute values become

|\sin x| = -\sin x

and

|x| = -x

, so

y = (-\sin x)^{-x}

. Use logarithmic differentiation to find

\frac{dy}{dx}

and evaluate at the given point.

Explanation:

Given

y = |\sin x|^{|x|}

and

x = -\frac{\pi}{6} < 0

, rewrite as

y = (-\sin x)^{-x}

Take natural logarithm:

\ln y = -x \ln(-\sin x)

Differentiate both sides with respect to

x

\frac{1}{y}\frac{dy}{dx} = -1 \cdot \ln(-\sin x) + (-x) \cdot \frac{1}{-\sin x} \cdot (-\cos x)

Simplify:

\frac{1}{y}\frac{dy}{dx} = -\ln(-\sin x) - x \cot x

Thus

\frac{dy}{dx} = y\left(-x \cot x - \ln(-\sin x)\right)

Substitute

y = (-\sin x)^{-x}

\frac{dy}{dx} = (-\sin x)^{-x} \left(-x \cot x - \ln(-\sin x)\right)

Now evaluate at

x = -\frac{\pi}{6}

-\sin\left(-\frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac{1}{2}

, so

(-\sin x)^{-x} = \left(\frac{1}{2}\right)^{\frac{\pi}{6}} = 2^{-\frac{\pi}{6}}

Next,

-x \cot x = -\left(-\frac{\pi}{6}\right) \cot\left(-\frac{\pi}{6}\right) = \frac{\pi}{6} \cdot (-\sqrt{3}) = -\frac{\pi\sqrt{3}}{6}

Also,

-\ln(-\sin x) = -\ln\left(\sin\frac{\pi}{6}\right) = -\ln\frac{1}{2} = \ln 2

Inside the parentheses:

-\frac{\pi\sqrt{3}}{6} + \ln 2 = \frac{6\ln 2 - \pi\sqrt{3}}{6}

Therefore,

\frac{dy}{dx}\big|_{x=-\pi/6} = 2^{-\frac{\pi}{6}} \cdot \frac{6\ln 2 - \pi\sqrt{3}}{6} = \frac{2^{-\frac{\pi}{6}}(6\ln 2 - \sqrt{3}\pi)}{6}

Answer:

\boxed{\frac{2^{-\frac{\pi }{6}}(6\ln 2-\sqrt{3}\pi )}{6}}

(Option A)

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