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Differentiation

Section: Mathematics

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Question : 75 of 79

Marks: +1, -0

Let

y

be an implicit function of

x

x^{2x}-2x^x\cot y-1=0

y'(1)

$1$
$\log2$
$-\log2$
$-1$

Solution:

\; x^{2x}-2x^x\cot y-1=0

\; \Rightarrow 2\cot y=x^x-x^{-x}

\; \Rightarrow 2\cot y=u-\frac{1}{u}

where

u=x^x

Differentiating both sides with respect to

x

,

we get

\Rightarrow -2\csc^2 y \; \frac{dy}{dx}

= \left(1+\frac{1}{u^2}\right) \; \frac{du}{dx}

where

u=x^x \Rightarrow \log u=x\log x

\; \Rightarrow \; \frac{1}{u} \frac{du}{dx}=1+\log x

\; \Rightarrow \; \frac{du}{dx}=x^x(1+\log x)

\; \therefore \; \text{ We get } -2\csc^2 y \; \frac{dy}{dx}

\; = (1+x^{-2x}) x^x(1+\log x)

\; \Rightarrow \; \frac{dy}{dx} = \frac{(x^x+x^{-x})(1+\log x)}{-2(1+\cot^2 y)}

.....(i)

Now when

x=1, x^{2x}-2x^x\cot y-1=0

gives

1-2\cot y-1=0

\Rightarrow \cot y=0

\therefore

From equation

(i)

, at

x=1

and

\cot y=0

, we get

y'(1)=\frac{(1+1)(1+0)}{-2(1+0)}=-1

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