Question:

The derivative of $cosec^{-1} \left(\frac{1}{2x\sqrt{1-x^{2}}}\right) w.r.t \sqrt{1-x^{2}}$ is

Updated On: Jul 7, 2022
  • $\frac{1}{\sqrt{1 - x^2}}$
  • $\frac{2}{x}$
  • $ - \frac{2}{x}$
  • $ - \frac{1}{ \sqrt{1 - x^2}}$
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The Correct Option is C

Solution and Explanation

Put $x = \sin\theta$ Let $ y = cosec^{-1} \left(\frac{1}{2x \sqrt{1-x^{2}}}\right)$ $ = cosec ^{-1} \left(\frac{1}{2\sin \theta \cos\theta}\right) = cosec^{-1} $ $\left(cosec 2 \theta\right) = 2 \theta$ $ \therefore \frac{dy}{dz} = \frac{dy/d\theta}{dz/d\theta} = \frac{2}{- \sin\theta} = -\frac{2}{x}$ $ z = \sqrt{1-x^{2}} = \sqrt{1-\sin^{2}\theta} = \cos\theta $
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Notes on Differentiability

Concepts Used:

Differentiability

Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability

(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.

3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.

(b) A function f(x) is differentiable in a closed interval [a, b] if it is

  • Differentiable at every point of interval (a, b)
  • Right derivative exists at x = a
  • Left derivative exists at x = b.