Advertisements
Advertisements
प्रश्न
Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2}, - \frac{1}{2 \sqrt{2}} \right)\] ?
Advertisements
उत्तर
\[\text { Let, u } = \sin^{- 1} \left( 4x\sqrt{1 - 4 x^2} \right)\]
\[ \text { put,} 2x = \cos\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \times \cos\theta\sqrt{1 - \cos^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\cos\theta \sin\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin 2\theta \right) . . . \left( i \right)\]
\[ \text {Let, v } = \sqrt{1 - 4 x^2} . . . \left( ii \right)\]
\[\text { Here }, \]
\[ x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[ \Rightarrow 2x \in \left( - 1, - \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow \theta \in \left( \frac{3\pi}{4}, \pi \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = \pi - 2\theta ......\left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \pi - \theta , \text{ if }\theta \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \right]\]
\[ \Rightarrow u = \pi - 2 \cos^{- 1} \left( 2x \right) ........\left[ \text { Since }, 2x = \cos\theta \right]\]
Differentiate it with respect to x,
\[\frac{du}{dx} = 0 - 2\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{4}{\sqrt{1 - 4 x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right), \]
\[\frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}}\]
\[\text { but }, x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[ \therefore \frac{dv}{dx} = \frac{- 4\left( - x \right)}{\sqrt{1 - 4 \left( - x \right)^2}}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{4x}{\sqrt{1 - 4 x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( iv \right)\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{4}{\sqrt{1 - 4 x^2}} \times \frac{\sqrt{1 - 4 x^2}}{4x}\]
\[ \therefore \frac{du}{dv} = \frac{1}{x}\]
APPEARS IN
संबंधित प्रश्न
Differentiate `2^(x^3)` ?
Differentiate \[e^{\sin^{- 1} 2x}\] ?
Differentiate \[3 e^{- 3x} \log \left( 1 + x \right)\] ?
Differentiate \[\cos \left( \log x \right)^2\] ?
If \[y = \log \sqrt{\frac{1 + \tan x}{1 - \tan x}}\] prove that \[\frac{dy}{dx} = \sec 2x\] ?
Differentiate \[\sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x \in R\] ?
Differentiate \[\tan^{- 1} \left( \frac{5 x}{1 - 6 x^2} \right), - \frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}\] ?
If \[y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0\] ,prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.
Find \[\frac{dy}{dx}\] in the following case \[\left( x^2 + y^2 \right)^2 = xy\] ?
If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\] ?
If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
Differentiate \[\left( \log x \right)^x\] ?
Differentiate \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\] ?
Find \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?
If \[y = \sin \left( x^x \right)\] prove that \[\frac{dy}{dx} = \cos \left( x^x \right) \cdot x^x \left( 1 + \log x \right)\] ?
If \[y = \log\frac{x^2 + x + 1}{x^2 - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{\sqrt{3} x}{1 - x^2} \right), \text{ find } \frac{dy}{dx} .\] ?
If \[x = a\sin2t\left( 1 + \cos2t \right) \text { and y } = b\cos2t\left( 1 - \cos2t \right)\] , show that at \[t = \frac{\pi}{4}, \frac{dy}{dx} = \frac{b}{a}\] ?
\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?
Differentiate x2 with respect to x3
Differentiate log (1 + x2) with respect to tan−1 x ?
Differentiate \[\left( \cos x \right)^{\sin x }\] with respect to \[\left( \sin x \right)^{\cos x }\]?
Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text { if }0 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\] with respect to \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right), \text { if } - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?
\[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cot^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right),\text { if }0 < x < 1\] ?
If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\] _________ .
If \[x^y = e^{x - y} ,\text{ then } \frac{dy}{dx}\] is __________ .
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
Find the second order derivatives of the following function x3 + tan x ?
Find the second order derivatives of the following function ex sin 5x ?
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
If y = 3 e2x + 2 e3x, prove that \[\frac{d^2 y}{d x^2} - 5\frac{dy}{dx} + 6y = 0\] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
If x = a cos nt − b sin nt and \[\frac{d^2 x}{dt} = \lambda x\] then find the value of λ ?
If x = 2at, y = at2, where a is a constant, then find \[\frac{d^2 y}{d x^2} \text { at }x = \frac{1}{2}\] ?
If x = t2, y = t3, then \[\frac{d^2 y}{d x^2} =\]
If x = 2 at, y = at2, where a is a constant, then \[\frac{d^2 y}{d x^2} \text { at x } = \frac{1}{2}\] is
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) w . r . t . \sin^{- 1} \frac{2x}{1 + x^2},\]tan-11+x2-1x w.r.t. sin-12x1+x2, if x ∈ (–1, 1) .
If x = a (1 + cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = \frac{- 1}{a}at \theta = \frac{\pi}{2}\]
f(x) = xx has a stationary point at ______.
