y-x-2-1-ln-1-x-1-1-x-2-dy-dx-

Question Number 108237 by Study last updated on 15/Aug/20

y = \sqrt{x^{2} + 1} - l n (\frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}})

\frac{d y}{d x} = ?

Answered by Dwaipayan Shikari last updated on 15/Aug/20

y = \sqrt{x^{2} + 1} - l o g (1 + \sqrt{x^{2} + 1}) + l o g x

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

Answered by john santu last updated on 15/Aug/20

You can't use 'macro parameter character #' in math mode

y = \sqrt{x^{2} + 1} - \ln (\frac{1 + \sqrt{1 + x^{2}}}{x})

\frac{d y}{d x} = \frac{x}{\sqrt{x^{2} + 1}} - (\frac{x}{1 + \sqrt{1 + x^{2}}}) (\frac{\frac{x^{2}}{\sqrt{1 + x^{2}}} - 1 - \sqrt{1 + x^{2}}}{x^{2}})

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

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

= \frac{x}{\sqrt{x^{2} + 1}} + \frac{1}{x \sqrt{x^{2} + 1}} = \frac{x^{2} + 1}{x \sqrt{x^{2} + 1}}

= \frac{\sqrt{x^{2} + 1}}{x}

Commented by bemath last updated on 15/Aug/20

t h e s i m p l e a n s w e r

Answered by mathmax by abdo last updated on 15/Aug/20

y (x) = \sqrt{x^{2} + 1} - \ln (\frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}}) \Rightarrow

y^{^{'}} (x) = \frac{x}{\sqrt{x^{2} + 1}} + \frac{{(\frac{1}{x} + \sqrt{1 + x^{- 2}})}^{^{'}}}{\frac{1}{x} + \sqrt{1 + x^{- 2}}} = \frac{x}{\sqrt{x^{2} + 1}} + \frac{- \frac{1}{x^{2}} + \frac{- {2 x}^{- 3}}{2 \sqrt{1 + x^{- 2}}}}{\frac{1}{x} + \sqrt{1 + x^{- 2}}}

= \frac{x}{\sqrt{1 + x^{2}}} - \frac{\frac{1}{x^{2}} + \frac{1}{x^{3} \sqrt{1 + x^{- 2}}}}{\frac{1}{x} + \sqrt{1 + x^{- 2}}} = \frac{x}{\sqrt{1 + x^{2}}} - \frac{1}{x^{3}} . \frac{x + \frac{1}{\sqrt{1 + x^{- 2}}}}{(\frac{1}{x} + \sqrt{1 + x^{- 2}})}

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