Find \(\frac {dy}{dx}:\) \(y=cos^{-1}(\frac {1-x^2}{1+x^2}),\ \ 0<x<1\)
Solution and Explanation
The given relationship is y = cos-1\((\frac {1-x^2}{1+x^2})\)
y = cos-1\((\frac {1-x^2}{1+x^2})\)
⇒cos y = \((\frac {1-x^2}{1+x^2})\)
⇒\(\frac {1-tan^2\frac y2}{1+tan^2\frac y2}\) = \(\frac {1-x^2}{1+x^2}\)
On comparing L.H.S. and R.H.S. of the above relationship, we obtain
tan \(\frac y2\) = x
Differentiating this relationship with respect to x,we obtain
sec2\(\frac y2\) . \(\frac {d}{dx}\)\((\frac y2)\) = \(\frac {d}{dx}\)(x)
⇒sec2\(\frac y2\) . \(\frac 12\)\(\frac {dy}{dx}\) = 1
⇒\(\frac {dy}{dx}\) = \(\frac {2}{sec^2\frac y2}\)
⇒\(\frac {dy}{dx}\)= \(\frac {2}{tan^2\frac y2}\)
∴\(\frac {dy}{dx}\) = \(\frac {1}{1+x^2}\)
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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.