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Question

Mathematics Question on Continuity and differentiability

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

A

−2(1+x2)-\frac {2}{(1+x^2)}

B

−1(1+x2)-\frac {1}{(1+x^2)}

C

2(1−x2)\frac {2}{(1-x^2)}

D

1(1+x2)\frac {1}{(1+x^2)}

Answer

−2(1+x2)-\frac {2}{(1+x^2)}

Explanation

Solution

Given: y = sec–1(x+x−1x−x−1)(\frac {x + x^{-1}}{x - x^{-1}})
y = sec–1(x+1xx−1x)(\frac {x + \frac{1}{x}}{x - \frac{1}{x}})
y = sec−1(x2+1x2−1)(\frac {x^2 +1}{x^2 -1})
Let x = cot θ
Then y = sec−1(cot2θ+1cot2θ−1)(\frac {cot^2θ +1}{cot^2θ -1})
We know that cos 2θ=1−tan2θ1+tan2θ\frac {1−tan^2θ}{1+tan^2θ}
y = sec-1(1cos 2θ)(\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 = −2(1+x2)-\frac {2}{(1+x^2)}
Therefore, the correct option is (A) −2(1+x2)-\frac {2}{(1+x^2)}