Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{ Let, u } = \sin^{- 1} \left( 4x\sqrt{1 - 4 x^2} \right)\]
\[ \text { put } 2x = \cos\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \times \cos\theta\sqrt{1 - \cos^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\cos\theta \sin\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin 2\theta \right) . . . \left( i \right)\]
\[ \text { Let, v }= \sqrt{1 - 4 x^2} . . . \left( ii \right)\]
\[\text{Here}, \]
\[ x \in \left( - \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right)\]
\[ \Rightarrow 2x \in \left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow \theta \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = \pi - 2\theta .....\left[ \text { Since }, \sin^{- 1} \left( sin \theta \right) = \pi - \theta , \text{ if }\theta \in \left( \frac{\pi}{2}, \pi \right) \right]\]
\[ \Rightarrow u = \pi - 2 \cos^{- 1} \left( 2x \right) \left[ \text { Since}, 2x = \cos\theta \right]\]
Differentiating it with respect to x,
\[\frac{du}{dx} = 0 - 2\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{4}{\sqrt{1 - 4 x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right)\]
\[\frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}}\]
\[\text { but,} x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[\frac{dv}{dx} = \frac{- 4\left( - x \right)}{\sqrt{1 - 4 \left( - x \right)^2}}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{4x}{\sqrt{1 - 4 x^2}} . . . \left( iv \right)\]
\[\text { Diferentiating equation } \left( ii \right) \text { with respect to x }, \]
\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\frac{d}{dx}\left( 1 - 4 x^2 \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\left( - 8x \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}} . . . \left( v \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( v \right)\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{4}{\sqrt{1 - 4 x^2}} \times \frac{\sqrt{1 - 4 x^2}}{- 4x}\]
\[ \therefore \frac{du}{dv} = - \frac{1}{x}\]
APPEARS IN
संबंधित प्रश्न
Differentiate \[e^{\sin} \sqrt{x}\] ?
Differentiate tan 5x° ?
Differentiate \[\sqrt{\frac{1 + \sin x}{1 - \sin x}}\] ?
Differentiate \[\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}\] ?
Differentiate \[\cos \left( \log x \right)^2\] ?
If \[y = \frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)\] , prove that \[\frac{ dy }{ dx } = 2 \text{cosec }2x \] ?
Differentiate \[\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?
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 \[\tan^{- 1} \left( \frac{a + x}{1 - ax} \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{x}{1 + 6 x^2} \right)\] ?
Differentiate the following with respect to x:
\[\cos^{- 1} \left( \sin x \right)\]
If \[y = \sin \left[ 2 \tan^{- 1} \left\{ \frac{\sqrt{1 - x}}{1 + x} \right\} \right], \text{ find } \frac{dy}{dx}\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\log x} + \cos^2 \left( \frac{\pi}{4} \right)\] ?
If \[y = x \sin y\] , prove that \[\frac{dy}{dx} = \frac{y}{x \left( 1 - x \cos y \right)}\] ?
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 \[xy = e^{x - y} , \text{ find } \frac{dy}{dx}\] ?
If \[y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + ... to \infty}}}\], prove that \[\left( 2 y - 1 \right) \frac{dy}{dx} = \frac{1}{x}\] ?
If \[y = \left( \cos x \right)^{\left( \cos x \right)^{\left( \cos x \right) . . . \infty}}\],prove that \[\frac{dy}{dx} = - \frac{y^2 \tan x}{\left( 1 - y \log \cos x \right)}\]?
Find \[\frac{dy}{dx}\] ,When \[x = e^\theta \left( \theta + \frac{1}{\theta} \right) \text{ and } y = e^{- \theta} \left( \theta - \frac{1}{\theta} \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{1 + ax}{1 - ax} \right)\] with respect to \[\sqrt{1 + a^2 x^2}\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - 1}{x + 1} \right)\] with respect to \[\sin^{- 1} \left( 3x - 4 x^3 \right), \text { if }- \frac{1}{2} < x < \frac{1}{2}\] ?
If \[- \frac{\pi}{2} < x < 0 \text{ and y } = \tan^{- 1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}, \text{ find } \frac{dy}{dx}\] ?
Find the second order derivatives of the following function x3 + tan x ?
If x = a sec θ, y = b tan θ, prove that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
If x = a cos θ, y = b sin θ, show that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
If x = a (θ + sin θ), y = a (1 + cos θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{a}{y^2}\] ?
If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0 ?
If x = 4z2 + 5, y = 6z2 + 7z + 3, find \[\frac{d^2 y}{d x^2}\] ?
If y = (cot−1 x)2, prove that y2(x2 + 1)2 + 2x (x2 + 1) y1 = 2 ?
\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text { and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?
\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]
If \[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\] then (1 − x)2 f '' (x) − xf(x) =
If y = (sin−1 x)2, then (1 − x2)y2 is equal to
If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?
