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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))

 
Answers (1)
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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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