Math, asked by cheel, 8 months ago

if y=x-√1+x^2, show that (1+x^2)(dy/dx)^2 = y^2​

Answers

Answered by pulakmath007
36

\displaystyle\huge\red{\underline{\underline{Solution}}}

GIVEN

y = x -  \sqrt{1 +  {x}^{2} }

TO PROVE

 \displaystyle \:  \: (1 +  {x}^{2} ) {( \frac{dy}{dx})}^{2}  =  {y}^{2}

CALCULATION

y = x -  \sqrt{1 +  {x}^{2} }

Differentiating both sides with respect to x we get

 \displaystyle \:  \frac{dy}{dx}  = 1 -  \frac{1}{2}  \times  \frac{2x}{ \sqrt{1 +  {x}^{2} } }

 \implies \:  \displaystyle \:  \frac{dy}{dx}  = 1 -  \frac{x}{ \sqrt{1 +  {x}^{2} } }

 \implies \:  \displaystyle \:  \frac{dy}{dx}  =   \frac{ \sqrt{1 -  {x}^{2} }  - x}{ \sqrt{1 +  {x}^{2} } }

 \implies \:  \displaystyle \:  \frac{dy}{dx}  =   \frac{  - (x - \sqrt{1 -  {x}^{2} }  )}{ \sqrt{1 +  {x}^{2} } }

 \implies \:  \displaystyle \:  \frac{dy}{dx}  =   \frac{  - y}{ \sqrt{1 +  {x}^{2} } }

Squaring both sides

 \implies \:  \displaystyle \: {( \frac{dy}{dx})}^{2}  =   \frac{  {y}^{2} }{{1 +  {x}^{2} } }

 \therefore \:  \displaystyle \:  \: (1 +  {x}^{2} ) {( \frac{dy}{dx})}^{2}  =  {y}^{2}

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