The derivative of $\cos^{-1} (2x^2 - 1)$ w.r.t $\cos^{-1} x$ is
- $2$
- $\frac{-1}{2\sqrt{1 -x^2}}$
- $\frac{2}{x}$
- $1 - x^2$
The Correct Option is A
Solution and Explanation
Now, $u=\cos ^{-1}\left(2 x^{2}-1\right)$
Put $x=\cos \theta$
$\therefore \, u=\cos ^{-1}\left(2 \cos ^{2} \theta-1\right)$
$=\cos ^{-1}(\cos 2 \theta)$
$=2 \theta=2 \cos ^{-1} \,x$
Again, $\frac{d u}{d v}=\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}$
$=\frac{\frac{d}{d x}\left(2 \cos ^{-1} x\right)}{\frac{d}{d x}\left(\cos ^{-1} x\right)}$
$=\frac{\left(-\frac{2}{\sqrt{1-x^{2}}}\right)}{\left(\frac{-1}{\sqrt{1-x^{2}}}\right)}=2$
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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.