\(Find\ \frac {dy}{dx}:\)
\(y=sin^{-1}(2x\sqrt {1-x^2},\ \frac {-1}{\sqrt 2}<x<\frac {1}{\sqrt2}\)
\(Find\ \frac {dy}{dx}:\)
\(y=sin^{-1}(2x\sqrt {1-x^2},\ \frac {-1}{\sqrt 2}<x<\frac {1}{\sqrt2}\)
Solution and Explanation
The given relationship is y=sin-1\((2x\sqrt {1-x^2}\)
y = sin-1\((2x\sqrt {1-x^2})\)
⇒siny = \((2x\sqrt {1-x^2}\)
Differentiating this relationship with respect to x, we obtain
cos y \(\frac {dy}{dx}\) = 2[x\(\frac {d}{dx}\)\((\sqrt {1-x^2})\) + \((\sqrt {1-x^2})\) \(\frac {dx}{dx}\)]
⇒\(\sqrt {1-sin^2y}\)\(\frac {dy}{dx}\) = 2[\(\frac x2\). -\(\frac {2x}{\sqrt{1-x^2}}\)+\(\sqrt{1-x^2}\)]
⇒\(\sqrt {1-(2x\sqrt {1-x2)^2}}\)\(\frac {dy}{dx}\) = 2\([\frac {-x^2+1-x^2}{√1-x^2}]\)
⇒\(\sqrt {1-4x^2(1-x^2)}\) \(\frac {dy}{dx}\)= 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒\(\sqrt {(1-2x^2)^2}\) \(\frac {dy}{dx}\)= 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒(1-2x2)\(\frac {dy}{dx}\) = 2\([\frac {1-2x^2}{√1-x^2}]\)
⇒\(\frac {dy}{dx}\) = \([\frac {2}{√1-x^2}]\)
Top Questions on Continuity and differentiability
- If the function
\[ f(x) = \begin{cases} \sin(2x), & \text{for } x > 0, \\ a + bx, & \text{for } x \leq 0, \end{cases} \]
where \( a \) and \( b \) are constants, is differentiable at \( x = 0 \), then \( a + b \) is equal to:- GATE BT - 2025
- Mathematics
- Continuity and differentiability
- If $$ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6, \text{ then } f(1) \text{ is equal to:} $$
- JEE Main - 2025
- Mathematics
- Continuity and differentiability
- Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice-differentiable function such that \( f(2) = 1 \). If \( F(x) = x f(x) \) for all \( x \in \mathbb{R} \), and the integrals \( \int_0^2 x F'(x) \, dx = 6 \) and \( \int_0^2 x^2 F''(x) \, dx = 40 \), then \( F'(2) + \int_0^2 F(x) \, dx \) is equal to:
- JEE Main - 2025
- Mathematics
- Continuity and differentiability
- For $a, b > 0$, let $ f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0, \\ \frac{x}{3}, & x = 0, \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{a x \sqrt{x}}}, & x > 0 \end{cases} $ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to:
- JEE Main - 2024
- Mathematics
- Continuity and differentiability
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is:
- IIITH UGEE - 2024
- Mathematics
- Continuity and differentiability
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.