Find \(\frac{dy}{dx} y=sin^{-1}(\frac{1-x^2}{1+x^2}),0<x<1\)
Solution and Explanation
The given relationship is y=sin-1(1-x2/1+x2)\(y=sin^{-1}(\frac{1-x^2}{1+x^2})\)
\(y=sin^{-1}(\frac{1-x^2}{1+x^2})\)
\(sin y=\frac{1-x^2}{1+x^2}\)
\(y=sin-1(\frac{1-x2}{1+x2})\)
\(⇒siny=(\frac{1-x2}{1+x2}) \)
\(⇒(1+x^2)siny=1-x^2\)
\(⇒(1+siny)x2^=1-siny\)
\(⇒x^2=\frac{1-siny}{1+siny}\)
\(⇒x^2=\frac{(cos\frac{y}{2}-sin\frac{y}{2})}{(cos\frac{y}{2}+sin\frac{y}{2})^2}\)
\(⇒x=\frac{cos\frac{y}{2}-sin\frac{y}{2}}{cos\frac{y}{2}+sin\frac{y}{2}}\)
\(⇒x=tan(\frac{π}{4}-\frac{y}{2})\)
Differentiating this relationship with respect to x, we obtain
\(\frac{d}{dx}(x)=\frac{d}{dx}.[tan(\frac{π}{4}-\frac{y}{2})]\)
\(⇒\frac{dy}{dx}=\frac{-2}{1+x^2}\)
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Notes on Continuity and 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.