If $ y = sec(tan^{-1}x) $ , then $ \frac{dy}{dx} $ at $ x = 1 $ is
- $ \frac{1}{2} $
- $ \frac{1}{\sqrt{2}} $
- $ \sqrt{2} $
- $ 1 $
The Correct Option is B
Solution and Explanation
$ = sec[sec^{-1} ( \sqrt{1 + x^2})]$
$[\because tan^{-1} \theta = sec^{-1} \sqrt{( 1 + \theta^2})]$
$\Rightarrow y = \sqrt{ 1 + x^2}$
On differentiating both sides w.r.t ?$x$?, we get
$\frac{dy}{dx} = \frac{1}{2} (1 + x^2) ^{-1/2} \frac{d}{dx} )( 1 + x^2)$
$ = \frac{1}{2} \frac{1}{\sqrt{1 + x^2}} \cdot 2x$
$\Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{(1 + x^2)}} $
$\Rightarrow (\frac{dy}{dx}) _{x = 1} = \frac{1}{\sqrt{(1+x^2)}} $
$= \frac{1}{\sqrt{2}}$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.