Advertisements
Advertisements
प्रश्न
Differentiate \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\] with respect to \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right), \text { if } - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?
उत्तर
\[\text { Let, u } = \tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\]
\[\text { Put x } = \sin\theta\]
\[ \Rightarrow \theta = \sin^{- 1} x\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\sin\theta}{\sqrt{1 - \sin^2 \theta}} \right) \]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\sin\theta}{\cos\theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \tan\theta \right) . . . \left( i \right)\]
\[\text { And }\]
\[\text { Let, v } = \sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)\]
\[ v = \sin^{- 1} \left( 2\sin\theta\sqrt{1 - \sin^2 \theta} \right)\]
\[ v = \sin^{- 1} \left( 2 \sin\theta\cos\theta \right)\]
\[ v = \sin^{- 1} \left( \sin2\theta \right) . . . \left( ii \right)\]
\[\text { Here,} \]
\[ - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\]
\[ \Rightarrow - \frac{1}{\sqrt{2}} < \sin\theta < \frac{1}{\sqrt{2}}\]
\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}\]
\[\text { So, from equation } \left( i \right), \]
\[u = \theta \left[ \text { Since,} \tan^{- 1} \left( \tan\theta \right) = \theta, \text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow u = \sin^{- 1} x\]
Differentiating it with respect to x,
\[\frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text{ from equation } \left( ii \right), \]
\[v = 2\theta \left[ \text { Since}, \sin^{- 1} \left( \sin\theta \right) = \theta, \text { if } \theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow v = 2 \sin^{- 1} x\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{2}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) \text {by}\left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \left( \frac{1}{\sqrt{1 - x^2}} \right)\left( \frac{\sqrt{1 - x^2}}{2} \right)\]
\[ \therefore \frac{du}{dv} = \frac{1}{2}\]
संबंधित प्रश्न
Prove that `y=(4sintheta)/(2+costheta)-theta `
Differentiate \[3^{e^x}\] ?
Differentiate \[3^{x^2 + 2x}\] ?
Differentiate \[3^{x \log x}\] ?
Differentiate \[\sqrt{\frac{1 - x^2}{1 + x^2}}\] ?
Differentiate \[\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}\] ?
Differentiate \[\frac{x^2 + 2}{\sqrt{\cos x}}\] ?
If \[y = \frac{x}{x + 2}\] , prove tha \[x\frac{dy}{dx} = \left( 1 - y \right) y\] ?
If \[y = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] .prove that \[\frac{dy}{dx} = 1 - y^2\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right), - \pi < x < \pi\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - a}{x + a} \right)\] ?
If \[y = \cos^{- 1} \left\{ \frac{2x - 3 \sqrt{1 - x^2}}{\sqrt{13}} \right\}, \text{ find } \frac{dy}{dx}\] ?
Find \[\frac{dy}{dx}\] in the following case \[x^{2/3} + y^{2/3} = a^{2/3}\] ?
If \[xy = 1\] prove that \[\frac{dy}{dx} + y^2 = 0\] ?
If \[x y^2 = 1,\] prove that \[2\frac{dy}{dx} + y^3 = 0\] ?
If \[\tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\] Prove that \[\frac{dy}{dx} = \frac{x}{y}\frac{\left( 1 - \tan a \right)}{\left( 1 + \tan a \right)}\] ?
If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?
If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?
Differentiate \[\left( 1 + \cos x \right)^x\] ?
Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?
Find \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?
If \[e^{x + y} - x = 0\] ,prove that \[\frac{dy}{dx} = \frac{1 - x}{x}\] ?
If \[x = e^{\cos 2 t} \text{ and y }= e^{\sin 2 t} ,\] prove that \[\frac{dy}{dx} = - \frac{y \log x}{x \log y}\] ?
Differentiate \[\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { if } 0 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text { if }0 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - 1}{x + 1} \right)\] with respect to \[\sin^{- 1} \left( 3x - 4 x^3 \right), \text { if }- \frac{1}{2} < x < \frac{1}{2}\] ?
If \[f'\left( x \right) = \sqrt{2 x^2 - 1} \text { and y } = f \left( x^2 \right)\] then find \[\frac{dy}{dx} \text { at } x = 1\] ?
If \[y = \sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] _____________ .
Find the second order derivatives of the following function tan−1 x ?
If y = (sin−1 x)2, prove that (1 − x2)
\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?
If y = ex (sin x + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?
If y log (1 + cos x), prove that \[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \cdot \frac{dy}{dx} = 0\] ?
If y = 3 e2x + 2 e3x, prove that \[\frac{d^2 y}{d x^2} - 5\frac{dy}{dx} + 6y = 0\] ?
If x = a cos nt − b sin nt, then \[\frac{d^2 x}{d t^2}\] is
If y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
If y = xn−1 log x then x2 y2 + (3 − 2n) xy1 is equal to
