The derivative of $tan ^{-1}[\frac{Sin x}{1+ Cos x}]$ with respect to $tan^{-1}[\frac{Cos x}{1+Sin x}]$ is
- 2
- -1
- 0
- -2
The Correct Option is B
Solution and Explanation
$u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right),$
$v=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$u =\tan ^{-1}\left(\frac{2 \sin x / 2 \cdot \cos x / 2}{1+2 \cos ^{2} x / 2-1}\right)$
$=\tan ^{-1}(\tan x / 2)=x / 2$
$v=\tan ^{-1}\left(\frac{\cos x}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)$
$=\tan ^{-1}\left(\frac{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}\right)$
$=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$
$=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$
$=\left[\tan ^{-1}\left(\tan ^{-1}\left(\frac{\pi}{4}-\frac{\pi}{2}\right)\right]=\left(\frac{\pi}{4}-\frac{x}{2}\right)\right.$
Now, $\frac{du}{dx}=\frac{1}{2}$ and $\frac{d v}{d x}=\frac{1}{2}$
$\Rightarrow \frac{d u}{d v}=\frac{d u}{d x} \times \frac{d x}{d v}$
$=\left(\frac{1}{2}\right) \times(-2)=-1$
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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.