Q

# Solve this problem - Limit , continuity and differentiability - JEE Main-18

Let $(\sin x)^{\cos x}$ then $\frac{\mathrm{d}y }{\mathrm{d} x}$ equals ?

• Option 1)

$(\cos x)^{-\sin x}$

• Option 2)

$(\cos x)(\sin x)^{-\cos x}$

• Option 3)

$(\sin x)^{-\sin x}$

• Option 4)

$(\sin x)^{\cos x}(\cos x \cdot \cot x - \sin x\cdot \log( \sin x))$

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As we have learnt,

Logarithmic differentiation -

When a function consists of the product of the quotient of a number  of functions we take logarithm.such as

$1.\:\:\: y={f_{1}(x)}^{f_{2}(x)}$

$2.\:\:\: y=f_{1}(x).f_{2}(x).f_{3}(x)......$

$2.\:\:\: y=\frac{f_{1}(x)\:f_{2}(x)\:f_{3}(x)....}{\phi_{1} (x)\:\phi_{2} (x)\:\phi _{3}(x)....}$

- wherein

For

$1.\:\:log\:y=f_{2}(x)log\:f_{1}(x)$

$2.\:\:log\:y=log\:f_{1}(x)+log\:f_{2}(x)+log\:f_{3}(x)+......$

$2.\:\:log\:y=[log\:f_{1}(x)+log\:f_{2}(x)+.....]-[log\:\phi _{1}(x)+log\:\phi _{2}(x)+....]$

$y = (\sin x)^{\cos x}$

Taking log on both sides,

$\log_{e}y = \log_{e}(\sin x)^{\cos x} \\*\Rightarrow \log_{e} y = \cos x\log_{e}\sin x$

Now diferenciating both sides,

$\frac{1}{y}\frac{dy}{dx} = \cos x \times \frac{\cos x}{\sin x} + \log_{e}(-\sin x) \\*\Rightarrow \frac{dy}{dx} = y(\cos x\cdot\cot x - (\sin x)\log_{e}(\sin x))$

Option 1)

$(\cos x)^{-\sin x}$

Option 2)

$(\cos x)(\sin x)^{-\cos x}$

Option 3)

$(\sin x)^{-\sin x}$

Option 4)

$(\sin x)^{\cos x}(\cos x \cdot \cot x - \sin x\cdot \log( \sin x))$

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