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  • Solve this problem - Limit , continuity and differentiability - JEE Main-16

An implicit function y such that \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\sin x}{y^2} then \frac{\mathrm{d} \sqrt{y}}{\mathrm{d} x} = ?

  • Option 1)

    \frac{1}{2}(\sin x)\cdot y^{\frac{-5}{2}}

  • Option 2)

    (\sin x)\cdot y^{\frac{3}{2}}

  • Option 3)

    (\sin x)\cdot \sqrt{y}

  • Option 4)

    (\sin x)\cdot y^{\frac{-5}{2}}

 

Answers (1)

best_answer

As we have learnt,

 

Derivative of implict function -

When  y is not expressible explicity in terms of  x  then we take, derivative of  y^{n}  as   ny^{n-1}\times \frac{dy}{dx}

- wherein

\frac{d}{dx}(y^{2})=2y.\frac{dy}{dx}


\frac{d}{dx}(y^{3})=3y^{2}.\frac{dy}{dx}

 

 \frac{\mathrm{d} }{\mathrm{d} x}\sqrt{y} = \frac{\mathrm{d} }{\mathrm{d} x}y^{\frac{1}{2}} = \frac{1}{2}(y)^{-\frac{1}{2}}\cdot\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2\sqrt{y}}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}\\* \Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\sqrt{y} =\frac{1}{2}(y)^{-\frac{1}{2}}\cdot \frac{\sin x}{y^{2}} = \frac{1}{2}(y)^{-\frac{5}{2}}\cdot \sin x

 


Option 1)

\frac{1}{2}(\sin x)\cdot y^{\frac{-5}{2}}

Option 2)

(\sin x)\cdot y^{\frac{3}{2}}

Option 3)

(\sin x)\cdot \sqrt{y}

Option 4)

(\sin x)\cdot y^{\frac{-5}{2}}

Posted by

Himanshu

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