\(Find\ \frac {dy}{dx}:\)
\(y=cos^{-1}(\frac {2x}{1+x^2}),\ -1<x<1\)
\(Find\ \frac {dy}{dx}:\)
\(y=cos^{-1}(\frac {2x}{1+x^2}),\ -1<x<1\)
Solution and Explanation
The given relationship is y = cos-1\((\frac {2x}{1+x^2})\)
y = cos-1\((\frac {2x}{1+x^2})\)
⇒cosy = \(\frac {2x}{1+x^2}\)
Differentiating this relationship with respect to x,we obtain
\(\frac {d}{dx}\)(cos y) = \(\frac {d}{dx}\)\((\frac {2x}{1+x^2})\)
⇒-sin y \(\frac {dy}{dx}\) = \(\frac {(1+x^2).\frac {d}{dx}(2x)-2x.\frac {d}{dx}(1+x^2)}{(1+x^2)^2}\)
⇒-\(\sqrt{1-cos^2y}\) \(\frac {dy}{dx}\) = \(\frac {(1+x^2).2-2x.2x}{(1+x^2)^2}\)
⇒\([\sqrt {1-(\frac {2x}{1+x^2})^2}\) \(\frac {dy}{dx}\) = -\([\frac {2(1-x^2)}{(1+x^2)^2}]\)
⇒\(\sqrt {\frac {(1-x^2)^2-4x^2}{(1+x^2)^2}}\) \(\frac {dy}{dx}\) = -\(\frac {2(1-x^2)}{(1+x^2)^2}\)
⇒\(\sqrt {\frac {(1-x^2)^2}{(1+x^2)^2}}\) \(\frac {dy}{dx}\) = -\(\frac {2(1-x^2)}{(1+x^2)^2}\)
⇒\(\frac {1-x^2}{1+x^2}\).\(\frac {dy}{dx}\)= -\(\frac {2(1-x^2)}{(1+x^2)^2}\)
⇒\(\frac {dy}{dx}\) = - \(\frac {2}{1+x^2}\)
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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.