MCQ
If $y = {\rm{sec}}\left( {{{\tan }^{ - 1}}x} \right)$ then $\frac{{dy}}{{dx}}$ at $x = 1$ is equal to :
- ✓$\frac{1}{{\sqrt 2 }}$
- B$\frac{1}{2}$
- C$1$
- D$\sqrt 2 $
✓
Answer
Correct option: A.
$\frac{1}{{\sqrt 2 }}$
a
$y=\sec \left(\tan ^{-1} x\right)$
$y=\sec \left(\tan ^{-1} x\right)$
Let $\quad \tan ^{-1} x=\theta$
${x=\tan \theta} $
${y=\sec \theta}$
${y=\sqrt{1+x^{2}}} $
${\frac{d y}{d x}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x} $
at ${x=1}$
$\frac{d y}{d x}=\frac{1}{\sqrt{2}}$
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