A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) 1
D) \[\sqrt{2}\]
Correct Answer: A
Solution :
\[y=sec(ta{{n}^{-1}}x)\] \[\Rightarrow \] \[\frac{dy}{dx}=sec(ta{{n}^{-1}}x)\cdot tan(ta{{n}^{-1}}x).\frac{1}{1+{{x}^{2}}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{x=1}}=\frac{\sqrt{2}}{1+1}=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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