If y=( x)^{x}+^{-1}{x}, then \;dy/dx

Correct Answer is : (sin⁡x)x[xcot⁡x+log⁡sin⁡x]+ 11−x× 12x( x)^{x}[x x+ x]+\;{1}{{1-x}} × \;{1}{2{x}}(sinx)x[xcotx+logsinx]+1−x1×2x1

( x)^{x}[ x+ x]+\;{1}{{1-x}} × \;{1}{2{x}}

( x)[x x+ x]+\;{1}{{1-x}} × \;{1}{2{x}}

( x)^{x}[x x+ x]+\;{1}{{1-x}} × \;{1}{2{x}}

Correct Answer is : (sin⁡x)x[xcot⁡x+log⁡sin⁡x]+ 11−x× 12x( x)^{x}[x x+ x]+\;{1}{{1-x}} × \;{1}{2{x}}(sinx)x[xcotx+logsinx]+1−x1×2x1

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MHT CET 2021 Math Solved Paper

Question : 81 of 100

Marks: +1, -0

y=(\sin x)^{x}+\sin^{-1}\sqrt{x}

\;\frac{dy}{dx}

$(\sin x)^{x}[\cot x+\log \sin x]+\;\frac{1}{\sqrt{1-x}} \times \;\frac{1}{2\sqrt{x}}$
$(\sin x)[x \cot x+\log \sin x]+\;\frac{1}{\sqrt{1-x}} \times \;\frac{1}{2\sqrt{x}}$
$(\sin x)^{x}[x \cot x+\log \sin x]+\;\frac{1}{\sqrt{1-x}} \times \;\frac{1}{2\sqrt{x}}$
None of the above

Solution:

Given,

y=(\sin x)^{x}+\sin^{-1}x

. . . (i)

Let

\;\; u=(\sin x)^{x}

. . . (ii)

Then, Eq. (i) becomes,

y=u+\sin^{-1}\sqrt{x}

. . . (iii)

On taking log both sides of Eq. (ii), we get

\log u=x \log \sin x

On differentiating both sides w.r.t.,

x

, we get

\;\frac{1}{u} \;\frac{du}{dx} \;\; =x \;\frac{d}{dx}(\log \sin x)+\log \sin x \;\frac{d}{dx}(x)

\Rightarrow \;\frac{du}{dx} \;\; =u\left[x \times \;\frac{1}{\sin x} \;\frac{d}{dx}(\sin x)+\log \sin x(1)\right]

\Rightarrow \;\; \;\frac{du}{dx} \;\; =(\sin x)^{x}\left[ \;\frac{x}{\sin x} \times \cos x+\log \sin x\right]

\Rightarrow \;\frac{du}{dx} \;\; =(\sin x)^{x}[x \cot x+\log \sin x]

. . . (iv)

On differentiating both sides of Eq. (iii) w.r.t.

x

, we get

\;\;\frac{dy}{dx}=\;\frac{du}{dx}+\;\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \;\frac{d}{dx}(\sqrt{x})

\therefore \;\; \;\frac{dy}{dx}=(\sin x)^{x}[x \cot x+\log \sin x]+\;\frac{1}{\sqrt{1-x}} \times \;\frac{1}{2\sqrt{x}}

[from Eq. (iv)]

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