\(Find\ \frac {dy}{dx}:\)
\(y=sin^{-1}(2x\sqrt {1-x^2},\ \frac {-1}{\sqrt 2}<x<\frac {1}{\sqrt2}\)
\(Find\ \frac {dy}{dx}:\)
\(y=sin^{-1}(2x\sqrt {1-x^2},\ \frac {-1}{\sqrt 2}<x<\frac {1}{\sqrt2}\)
Solution and Explanation
The given relationship is y=sin-1\((2x\sqrt {1-x^2}\)
y = sin-1\((2x\sqrt {1-x^2})\)
⇒siny = \((2x\sqrt {1-x^2}\)
Differentiating this relationship with respect to x, we obtain
cos y \(\frac {dy}{dx}\) = 2[x\(\frac {d}{dx}\)\((\sqrt {1-x^2})\) + \((\sqrt {1-x^2})\) \(\frac {dx}{dx}\)]
⇒\(\sqrt {1-sin^2y}\)\(\frac {dy}{dx}\) = 2[\(\frac x2\). -\(\frac {2x}{\sqrt{1-x^2}}\)+\(\sqrt{1-x^2}\)]
⇒\(\sqrt {1-(2x\sqrt {1-x2)^2}}\)\(\frac {dy}{dx}\) = 2\([\frac {-x^2+1-x^2}{√1-x^2}]\)
⇒\(\sqrt {1-4x^2(1-x^2)}\) \(\frac {dy}{dx}\)= 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒\(\sqrt {(1-2x^2)^2}\) \(\frac {dy}{dx}\)= 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒(1-2x2)\(\frac {dy}{dx}\) = 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒\(\frac {dy}{dx}\) = \([\frac {2}{√1-x^2}]\)
Top Questions on Continuity and differentiability
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(i) Find \(f'(x)\) for \(0<x>3\).
(ii) Find \(f'(4)\).
(iii)(a) Test for continuity of \(f(x)\) at \(x=3\).
OR
(iii)(b) Test for differentiability of \(f(x)\) at \(x=3\).
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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.