If tan^{-1}\left [ \frac{Sinx+Cosx}{Cosx-Sinx} \right ], then \frac{dy}{dx} is equal to

  • Option 1)

    \frac{1}{2}

  • Option 2)

    \frac{\pi }{4}

  • Option 3)

    0

  • Option 4)

    1

 

Answers (1)

As we discussed in concept

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

 

 If y\:=\:tan^{-1}\:(\frac{cosx+sinx}{cosx-sinx})

        = tan^{-1}\:(\frac{1+tanx}{1-tanx})

       = tan^{-1}\:tan(\frac{\pi}{4}+x)

       = \frac{\pi}{4}+x

\frac{dy}{dx}=0+1=1


Option 1)

\frac{1}{2}

This option is incorrect.

Option 2)

\frac{\pi }{4}

This option is incorrect.

Option 3)

0

This option is incorrect.

Option 4)

1

This option is correct.

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