Limit Continuity And Differentiability Ques 4
- If $y=\sec \left(\tan ^{-1} x\right)$, then $\frac{d y}{d x}$ at $x=1$ is equal to
(2013)
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) $1$
(d) $\sqrt{2}$
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Answer:
Correct Answer: 4.(a)
Solution: (a) Given, $y=\sec \left(\tan ^{-1} x\right)$
$\text { Let } \tan ^{-1} x =\theta $
$\Rightarrow \quad x =\tan \theta $
$ \therefore \quad y =\sec \theta=\sqrt{1+x^2}$
On differentiating w.r.t. $x$, we get
$ \frac{d y}{d x}=\frac{1}{2 \sqrt{1+x^2}} \cdot 2 x $
At $x=1, \quad \frac{d y}{d x}=\frac{1}{\sqrt{2}}$
BETA
