Question

If y = sec^–1\((\frac {x + x^{-1}}{x - x^{-1}})\), then \(\frac {dy}{dx}\) =?

• A. \(-\frac {2}{(1+x^2)}\)
• B. \(-\frac {1}{(1+x^2)}\)
• C. \(\frac {2}{(1-x^2)}\)
• D. \(\frac {1}{(1+x^2)}\)

Answer 1

The Correct Option is A

Given: y = sec^–1\((\frac {x + x^{-1}}{x - x^{-1}})\)
y = sec^–1\((\frac {x + \frac{1}{x}}{x - \frac{1}{x}})\)
y = sec⁻¹\((\frac {x^2 +1}{x^2 -1})\)
Let x = cot θ
Then y = sec⁻¹\((\frac {cot^2θ +1}{cot^2θ -1})\)
We know  that cos 2θ=\(\frac {1−tan^2θ}{1+tan^2θ}\)
y = sec^-1\((\frac {1}{cos \ 2θ})\)
y = sec^-1(sec 2θ)
y = 2θ
put θ = cot^-1 x
y = 2 cot^-1 x
Differentiate both side
dy/dx = \(-\frac {2}{(1+x^2)}\)
Therefore, the correct option is (A) \(-\frac {2}{(1+x^2)}\)