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Let y be an implicit function of  x defined by x^{2x}-2x^{x}\cot y-1= 0

then {y}'\left ( 1 \right ) equals

  • Option 1)

    1

  • Option 2)

    \log 2

  • Option 3)

    -\log 2

  • Option 4)

    0

 

Answers (1)

best_answer

As we learnt in 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

 

 x^{2x}-2x^xcoty -1 =0

2x^{2x}(1+logx)-2\left [ coty.x^x(1+logx)+x^x -cosec^2y.y' \right ] =0

\Rightarrow x^{2x}(1+logx)- coty.x^x(1+logx)+x^x.y' cosec^2y =0

Put x=1

1(1+log\:1)- cot\:y.1(1+log1)+ 1.y' cosec^2y =0

\frac{1-cot\:y}{cosec^2y} =\frac{dy}{dx}                        at   x=1, cot\:y=0, \:cosec^2y=1

\Rightarrow \frac{1-0}{1} =\frac{dy}{dx}=1

Correct optoin is 1


Option 1)

1

Correct

Option 2)

\log 2

Incorrect

Option 3)

-\log 2

Incorrect

Option 4)

0

Incorrect

Posted by

Aadil

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