Advertisements
Advertisements
Question
If \[y = \sqrt{x + 1} + \sqrt{x - 1}\] , prove that \[\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y\] ?
Solution
\[\text{We have }, y = \sqrt{x + 1} + \sqrt{x - 1}\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x + 1} \right) + \frac{d}{dx}\left( \sqrt{x - 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} + \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2} \]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{\sqrt{x + 1}} + \frac{1}{\sqrt{x - 1}} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{\sqrt{x - 1} + \sqrt{x + 1}}{\left( \sqrt{x + 1} \right)\left( \sqrt{x - 1} \right)} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{y}{\left( \sqrt{x^2 - 1} \right)} \right)\]
\[ \Rightarrow \left( \sqrt{x^2 - 1} \right)\frac{d y}{d x} = \frac{1}{2}y\]
\[\text{Hence proved }\]
APPEARS IN
RELATED QUESTIONS
Differentiate the following functions from first principles e3x.
Differentiate \[\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}\] ?
Differentiate \[\sqrt{\frac{1 + \sin x}{1 - \sin x}}\] ?
Differentiate \[\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}\] ?
Differentiate \[\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}\] ?
Differentiate \[\cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}, \frac{1}{\sqrt{2}} < x < 1\] ?
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 \[y = \cot^{- 1} \left\{ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right\}\], show that \[\frac{dy}{dx}\] is independent of x. ?
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 \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
If \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\] \[\frac{dy}{dx} \] ?
Find \[\frac{dy}{dx}\] in the following case \[x^5 + y^5 = 5 xy\] ?
If \[\sqrt{1 - x^2} + \sqrt{1 - y^2} = a \left( x - y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - x^2}\] ?
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 \[x^{1/x}\] with respect to x.
If \[e^x + e^y = e^{x + y}\] , prove that
\[\frac{dy}{dx} + e^{y - x} = 0\] ?
If \[xy \log \left( x + y \right) = 1\] , prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
If \[x = e^{\cos 2 t} \text{ and y }= e^{\sin 2 t} ,\] prove that \[\frac{dy}{dx} = - \frac{y \log x}{x \log y}\] ?
If \[x = a \left( \theta - \sin \theta \right) and, y = a \left( 1 + \cos \theta \right), \text { find } \frac{dy}{dx} \text{ at }\theta = \frac{\pi}{3} \] ?
Differentiate \[\tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\] with respect to \[\sqrt{1 - x^2},\text {if} - 1 < x < 1\] ?
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] write the value of \[\frac{dy}{dx}\text { for } x > 1\] ?
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 \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
If y = log (sin x), prove that \[\frac{d^3 y}{d x^3} = 2 \cos \ x \ {cosec}^3 x\] ?
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?
If \[y = \left[ \log \left( x + \sqrt{x^2 + 1} \right) \right]^2\] show that \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\] ?
If \[y = e^{a \cos^{- 1}} x\] ,prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - a^2 y = 0\] ?
If y = sin (log x), prove that \[x^2 \frac{d^2 y}{d x^2} + x\frac{dy}{dx} + y = 0\] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
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 y = a xn + 1 + bx−n and \[x^2 \frac{d^2 y}{d x^2} = \lambda y\] then write 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 y = x + ex, find \[\frac{d^2 x}{d y^2}\] ?
If y = |x − x2|, then find \[\frac{d^2 y}{d x^2}\] ?
If x = t2, y = t3, then \[\frac{d^2 y}{d x^2} =\]
If logy = tan–1 x, then show that `(1+x^2) (d^2y)/(dx^2) + (2x - 1) dy/dx = 0 .`
If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .
\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]
\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]
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?
