Question

If y= [x+1/x](x-1/x+1),then dy/dx

Answer 1

$\underline{\textbf{\large{Answer:}}}$

$\boxed{\sf \frac{dy}{dx}= \frac{x - 1}{ {x}^{3} + {x}^{2} }}$

$\underline{\textbf{\large{Explanation:}}}$

$\sf y =( \frac{x + 1}{x} )( \frac{x - 1}{x + 1} )$

$\underline {\textbf{Differentiation wrt x}}$

$\sf \frac{dy}{dx} = ( \frac{x + 1}{x} )( \frac{(x + 1)(1 - 0) - ((x - 1)(1 + 0))}{ {(x + 1)}^{2} } ) \\ + \sf (( \frac{x - 1}{x + 1} )( \frac{x(1 + 0) - (x + 1)(1)}{ {x}^{2} } ))$

$\sf =( \frac{x + 1}{x} )( \frac{(x + 1) - (x - 1)}{ {(x + 1)}^{2} } ) \\ +\sf ( \frac{x - 1}{x + 1} )( \frac{x - (x + 1)}{( {x}^{2}) } )$

$\sf = ( \frac{x + 1}{x} )( \frac{x + 1 - x + 1}{ {(x + 1)}^{2} } ) + ( \frac{x - 1}{x + 1} )( \frac{x - x - 1}{ {x}^{2} } )$

$\sf = \frac{2}{x(x + 1)} +( \frac{ - (x - 1)}{ {x}^{2}(x + 1) } )$

$\sf = \frac{2 {x}^{2}(x + 1) + ( - (x - 1)( {x}^{2} + x)) }{x(x + 1)( {x}^{2}(x + 1)) }$

$\sf = \frac{2 {x}^{2} (x + 1) + ( - (x - 1)x(x + 1))}{x(x + 1)( {x}^{2} (x + 1))}$

$\sf = \frac{2 {x}^{2} + ( - x(x - 1)) }{ {x}^{3}(x + 1) }$

$\sf = \frac{2 {x}^{2} - {x}^{2} - x}{ {x}^{3} (x + 1)}$

$\sf = \frac{2x - x - 1}{ {x}^{3} + {x}^{2} }$

$\boxed{ \sf\frac{dy}{dx}= \frac{x - 1}{ {x}^{3} + {x}^{2} }}$