Math, asked by talpadadilip417, 19 days ago


   \pink{\boxed{ \tt \red{Diff. W. R  \: to \:  x: \boxed{ \tt \sqrt{ \dfrac{x + 1}{x - 1} } }}}}
Answer with proper explanation.

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Answers

Answered by Anonymous
10

\sf\red{Solution:-}

Answer in the above attachment.

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Additional information:-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered} \\ \end{gathered}

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Answered by gops2k4
1

Answer:

Step-by-step explanation:

\frac{d}{dx} [\sqrt{\frac{x+1}{x-1} } ]\\= \frac{1}{2\sqrt{\frac{x+1}{x-1} }} * \frac{1(x-1)-(x+1)1}{(x-1^){2} } \\=\frac{-1}{\sqrt{\frac{x+1}{x-1} }*(x-1)^{2}  } \\=\frac{-\sqrt{x-1} }{\sqrt{x+1}(x-1)^{2}  }\\=\frac{-(x-1)^{\frac{-3}{2} } }{\sqrt{x+1} }\\=\frac{-1}{\sqrt{x+1}\sqrt{x-1}(x-1)  }  \\=\frac{-1}{\sqrt{x^{2} -1}(x-1) }

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