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  • Can someone explain - Limit , continuity and differentiability - JEE Main-12

Let y = \sin(x + y) then \frac{\mathrm{d} y}{\mathrm{d} x} = ?

  • Option 1)

    \frac{\cos(x+y)}{1 + \cos(x+y)}

  • Option 2)

    \frac{\cos(x+y)}{\cos(x+y)-1}

  • Option 3)

    \frac{\cos(x+y)}{1 - \cos(x+y)}

  • Option 4)

    -\frac{\cos(x+y)}{1 + \cos(x+y)}

 

Answers (1)

best_answer

As we have learnt,

 

Derivative of implict function -

When  y  is given in any function then we find derivative of function first then find derivative of  y and collect the terms containing  

dy / dx  on left side and find  dy / dx in terms of  x & y

- wherein

ex:

Let \:\:y=siny

\frac{dy}{dx}=cosy\times\frac{dy}{dx}

\frac{dy}{dx}(1-cosy)=0

ex:

Let\:\:y=sin(xy)

\frac{dy}{dx}=cos(xy)\times(1.y+\frac{xdy}{dx})

\frac{dy}{dx}(1-x\:cos(xy))=y\:cos(xy)

\therefore \:\frac{dy}{dx}=\frac{y\:cos(xy)}{1-x\:cos(xy)}

 

 Defferenciating both sides we have,

\frac{\mathrm{d} y}{\mathrm{d} x} = \left (\cos(x +y) \right )\left (1 + \frac{\mathrm{d} y}{\mathrm{d} x} \right ) \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \cos(x+y) + \frac{\mathrm{d} y}{\mathrm{d} x}\cdot\cos(x+y)

\\*\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}(1 - \cos(x +y)) = \cos(x+y) \\* \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\cos(x+y)}{1 - \cos(x+y)}

 


Option 1)

\frac{\cos(x+y)}{1 + \cos(x+y)}

Option 2)

\frac{\cos(x+y)}{\cos(x+y)-1}

Option 3)

\frac{\cos(x+y)}{1 - \cos(x+y)}

Option 4)

-\frac{\cos(x+y)}{1 + \cos(x+y)}

Posted by

Himanshu

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