\(\text {Find} \ \frac {dy}{dx}:\) \(y=sin^{-1}(\frac {2x}{1+x^2})\)
Solution and Explanation
Differentiating this relationship with respect to x, we obtain
\(\frac {d}{dx}\)(sin y) = \(\frac {d}{dx}\) \((\frac {2x}{1+x^2})\)
\(\implies\)cos y \(\frac {dy}{dx}\) = \(\frac {d}{dx}\)\((\frac {2x}{1+x^2})\) …….... (1)
The function, \((\frac {2x}{1+x^2})\), is of the form of \(\frac uv\).
Therefore, by quotient rule, we obtain
\(\frac {d}{dx}\) \((\frac {2x}{1+x^2})\)=(1+x2) . \(\frac {d}{dx}\)(2x) -2x . \(\frac {d}{dx}\)\((\frac {1+x^2}{1+x^2})\)
= \(\frac {(1+x^2) . 2-2x(0+2x)}{(1+x^2)^2}\)
= \(\frac {2+2x^2-4x^2}{(1+x^2)^2}\)
= \(\frac {2(1-x^2)}{(1+x^2)^2}\) ……..… (2)
Also, sin y = \(\frac {2x}{1+x^2}\)
\(\implies\)cos y = \(\sqrt {1-sin^2y}\) = \(\sqrt {1- (\frac {2x}{1+x^2})^2}\) = \(\sqrt\frac { (1+x^2)^2-4x^2}{(1+x^2)^2}\)
= \(\sqrt {\frac {(1-x^2)^2}{(1+x^2)^2}}\)
= \( \frac {1-x^2}{1+x^2}\)
From (1), (2) and (3) we obtain
\( \frac {1-x^2}{1+x^2}\) . \(\frac {dy}{dx}\) = \( \frac {2(1-x^2)}{(1+x^2)}\)
\(\implies\) \(\frac {dy}{dx}\) = \(\frac {2}{1+x^2}\)
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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.