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 tan2 x ?
Differentiate \[\sqrt{\frac{1 + x}{1 - x}}\] ?
Differentiate \[e^{\tan 3 x} \] ?
Differentiate \[\log \left( x + \sqrt{x^2 + 1} \right)\] ?
Differentiate \[e^{ax} \sec x \tan 2x\] ?
If \[y = \frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)\] , prove that \[\frac{ dy }{ dx } = 2 \text{cosec }2x \] ?
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
Differentiate \[\cos^{- 1} \left\{ \sqrt{\frac{1 + x}{2}} \right\}, - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\] ?
Find \[\frac{dy}{dx}\] in the following case \[xy = c^2\] ?
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 \[e^{x \log x}\] ?
Differentiate \[\left( \sin^{- 1} x \right)^x\] ?
Differentiate \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\] ?
Find \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x}\] ?
If \[x^{16} y^9 = \left( x^2 + y \right)^{17}\] ,prove that \[x\frac{dy}{dx} = 2 y\] ?
If \[x^x + y^x = 1\], prove that \[\frac{dy}{dx} = - \left\{ \frac{x^x \left( 1 + \log x \right) + y^x \cdot \log y}{x \cdot y^\left( x - 1 \right)} \right\}\] ?
If \[y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . to \infty}}}\] , prove that \[\frac{dy}{dx} = \frac{\sec^2 x}{2 y - 1}\] ?
If \[y = \left( \cos x \right)^{\left( \cos x \right)^{\left( \cos x \right) . . . \infty}}\],prove that \[\frac{dy}{dx} = - \frac{y^2 \tan x}{\left( 1 - y \log \cos x \right)}\]?
Differentiate (log x)x with respect to log x ?
If f (x) = loge (loge x), then write the value of `f' (e)` ?
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] write the value of \[\frac{dy}{dx}\text { for } x > 1\] ?
The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is ___________ .
Differential coefficient of sec(tan−1 x) is ______.
Find the second order derivatives of the following function ex sin 5x ?
If \[y = \frac{\log x}{x}\] show that \[\frac{d^2 y}{d x^2} = \frac{2 \log x - 3}{x^3}\] ?
If y = ex cos x, prove that \[\frac{d^2 y}{d x^2} = 2 e^x \cos \left( x + \frac{\pi}{2} \right)\] ?
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(1 − cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{1}{a}\text { at } \theta = \frac{\pi}{2}\] ?
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
If x = a cos nt − b sin nt and \[\frac{d^2 x}{dt} = \lambda x\] then find the value of λ ?
If \[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}\] .....to ∞, then write \[\frac{d^2 y}{d x^2}\] in terms of y ?
If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to
If \[y = \tan^{- 1} \left\{ \frac{\log_e \left( e/ x^2 \right)}{\log_e \left( e x^2 \right)} \right\} + \tan^{- 1} \left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)\], then \[\frac{d^2 y}{d x^2} =\]
If \[y = \frac{ax + b}{x^2 + c}\] then (2xy1 + y)y3 =
If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?
