1.

If y=x+√(x²-1), then y-x(dy/dx) = ???

Answer»

If y=X+√(x²-1), then y-x(dy/dx) = ???

\textbf{Given:}

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

\textbf{To <klux>FIND</klux>:}

y-x\dfrac{dy}{dx}

\textbf{Solution:}

\text{Consider,}

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

\text{Differentiate with respect to 'x'}

\dfrac{dy}{dx}=1+\dfrac{<klux>2X</klux>}{2\sqrt{x^2-1}}

\dfrac{dy}{dx}=1+\dfrac{x}{\sqrt{x^2-1}}

\text{Now,}

y-x\dfrac{dy}{dx}

=(x+\sqrt{x^2-1})-x(1+\dfrac{x}{\sqrt{x^2-1}})

=\sqrt{x^2-1}-\dfrac{x^2}{\sqrt{x^2-1}}

=\dfrac{(x^2-1)-x^2}{\sqrt{x^2-1}}

=\dfrac{-1}{\sqrt{x^2-1}}

\implies\boxed{\bf\,y-x\dfrac{dy}{dx}=\dfrac{-1}{\sqrt{x^2-1}}}

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