Advertisements
Advertisements
प्रश्न
Differentiate \[\tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\] with respect to \[\sqrt{1 - x^2},\text {if} - 1 < x < 1\] ?
Advertisements
उत्तर
\[\text { Let, u } = \tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\]
\[\text { Put x }= \tan\theta\]
\[ \Rightarrow \theta = \tan^{- 1} x\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{1 - \tan\theta}{1 + \tan\theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} - \theta \right) \right] . . . \left( i \right)\]
\[\text { Here,} \]
\[ - 1 < x < 1\]
\[ \Rightarrow - 1 < \tan\theta < 1\]
\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}\]
\[ \Rightarrow \frac{\pi}{4} > - \theta > \frac{\pi}{4}\]
\[ \Rightarrow - \frac{\pi}{4} < - \theta < \frac{\pi}{4}\]
\[ \Rightarrow 0 < \frac{\pi}{4} - \theta < \frac{\pi}{2}\]
\[\text { So, from equation } \left( i \right), \]
\[u = \frac{\pi}{4} - \theta \left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow u = \frac{\pi}{4} - \tan^{- 1} x\]
Differentiating it with respect to x,
\[\frac{du}{dx} = 0 - \left( \frac{1}{1 + x^2} \right)\]
\[ \Rightarrow \frac{du}{dx} = - \frac{1}{1 + x^2} . . . \left( ii \right)\]
\[\text {And let, v } = \sqrt{1 - x^2}\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}} \times \frac{d}{dx}\left( 1 - x^2 \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}}\left( - 2x \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{- x}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text { Dividing equation }\left( ii \right) by \left( iii \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = - \frac{1}{1 + x^2} \times \frac{\sqrt{1 - x^2}}{- x}\]
\[ \therefore \frac{du}{dv} = \frac{\sqrt{1 - x^2}}{x\left( 1 + x^2 \right)}\]
APPEARS IN
संबंधित प्रश्न
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60º.
Differentiate \[e^{3 x} \cos 2x\] ?
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Differentiate \[e^{ax} \sec x \tan 2x\] ?
If \[y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]prove that \[\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}\] ?
If \[y = \frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)\] , prove that \[\frac{ dy }{ dx } = 2 \text{cosec }2x \] ?
Find \[\frac{dy}{dx}\] in the following case \[\tan^{- 1} \left( x^2 + y^2 \right) = a\] ?
If \[xy = 1\] prove that \[\frac{dy}{dx} + y^2 = 0\] ?
If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that \[\frac{dy}{dx} = \frac{y}{x}\] ?
Differentiate \[\left( \log x \right)^x\] ?
Differentiate \[\sin \left( x^x \right)\] ?
Differentiate \[\left( \cos x \right)^x + \left( \sin x \right)^{1/x}\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x}\] ?
If \[e^x + e^y = e^{x + y}\] , prove that
\[\frac{dy}{dx} + e^{y - x} = 0\] ?
If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] , prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?
Find \[\frac{dy}{dx}\], When \[x = a \left( \theta + \sin \theta \right) \text{ and } y = a \left( 1 - \cos \theta \right)\] ?
Write the derivative of sinx with respect to cos x ?
Differentiate log (1 + x2) with respect to tan−1 x ?
Differentiate \[\tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\] with respect to \[\sec^{- 1} x\] ?
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 \[y = \sin^{- 1} \left( \sin x \right), - \frac{\pi}{2} \leq x \leq \frac{\pi}{2}\] ,Then, write the value of \[\frac{dy}{dx} \text{ for } x \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \] ?
If \[f\left( 0 \right) = f\left( 1 \right) = 0, f'\left( 1 \right) = 2 \text { and y } = f \left( e^x \right) e^{f \left( x \right)}\] write the value of \[\frac{dy}{dx} \text{ at x } = 0\] ?
If \[y = \sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text{ find } \frac{dy}{dx}\] ?
If f (x) = logx2 (log x), the `f' (x)` at x = e is ____________ .
If \[y = \left( 1 + \frac{1}{x} \right)^x , \text{then} \frac{dy}{dx} =\] ____________.
Given \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .
If \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .
If \[y = \log \sqrt{\tan x}\] then the value of \[\frac{dy}{dx}\text { at }x = \frac{\pi}{4}\] is given by __________ .
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?
Find \[\frac{d^2 y}{d x^2}\] where \[y = \log \left( \frac{x^2}{e^2} \right)\] ?
If x = 4z2 + 5, y = 6z2 + 7z + 3, find \[\frac{d^2 y}{d x^2}\] ?
If y = 3 e2x + 2 e3x, prove that \[\frac{d^2 y}{d x^2} - 5\frac{dy}{dx} + 6y = 0\] ?
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 \[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}\] .....to ∞, then write \[\frac{d^2 y}{d x^2}\] in terms of y ?
\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]
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
If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?
If p, q, r, s are real number and pr = 2(q + s) then for the equation x2 + px + q = 0 and x2 + rx + s = 0 which of the following statement is true?
