Advertisements
Advertisements
प्रश्न
If \[\sqrt{y + x} + \sqrt{y - x} = c, \text {show that } \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\] ?
Advertisements
उत्तर
\[\text{ Here,} \] \[ \sqrt{y + x} + \sqrt{y - x} = c\]
Differentiating with respect to x,
\[\Rightarrow \frac{d}{dx}\left( \sqrt{y + x} \right) + \frac{d}{dx}\sqrt{y - x} = \frac{d}{dx}\left( c \right)\]
\[ \Rightarrow \frac{1}{2\sqrt{y + x}}\frac{d}{dx}\left( y + x \right) + \frac{1}{2\sqrt{y - x}}\frac{d}{dx}\left( y - x \right) = 0 \]
\[ \Rightarrow \frac{1}{2\sqrt{y + x}}\left( \frac{dy}{dx} + 1 \right) + \frac{1}{2\sqrt{y - x}}\left( \frac{dy}{dx} - 1 \right) = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1}{2\sqrt{y + x}} \right) + \frac{dy}{dx}\left( \frac{1}{2\sqrt{y - x}} \right) = \frac{1}{2\sqrt{y - x}} - \frac{1}{2\sqrt{y + x}}\]
\[ \Rightarrow \frac{dy}{dx} \times \frac{1}{2}\left[ \frac{1}{\sqrt{y + x}} + \frac{1}{\sqrt{y - x}} \right] = \frac{1}{2}\left[ \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y - x}\sqrt{y + x}} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sqrt{y - x} + \sqrt{y + x}}{\sqrt{y + x}\sqrt{y - x}} \right] = \left[ \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y - x}\sqrt{y + x}} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y + x} + \sqrt{y - x}} \times \frac{\sqrt{y + x} - \sqrt{y - x}}{\sqrt{y + x} - \sqrt{y - x}} \left[ \text{ rationalizing the denominator } \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( y + x \right) + \left( y - x \right) - 2\sqrt{y + x}\sqrt{y - x}}{y + x - y + x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2y - 2\sqrt{y^2 - x^2}}{2x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2y}{2x} - \frac{2\sqrt{y^2 - x^2}}{2x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2 - x^2}{x^2}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\]
APPEARS IN
संबंधित प्रश्न
If y = xx, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`
Differentiate the following functions from first principles eax+b.
Differentiate `2^(x^3)` ?
Differentiate \[e^\sqrt{\cot x}\] ?
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Differentiate \[\frac{x^2 + 2}{\sqrt{\cos x}}\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
Differentiate \[\sin^{- 1} \left( 2 x^2 - 1 \right), 0 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
Differentiate \[\cos^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] ?
Differentiate \[\cos^{- 1} \left( \frac{1 - x^{2n}}{1 + x^{2n}} \right), < x < \infty\] ?
Differentiate \[\tan^{- 1} \left( \frac{a + x}{1 - ax} \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{a + bx}{b - ax} \right)\] ?
If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?
Find \[\frac{dy}{dx}\] in the following case \[\left( x + y \right)^2 = 2axy\] ?
Find \[\frac{dy}{dx}\] in the following case \[\sin xy + \cos \left( x + y \right) = 1\] ?
If \[y = \left\{ \log_{\cos x} \sin x \right\} \left\{ \log_{\sin x} \cos x \right\}^{- 1} + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right), \text{ find } \frac{dy}{dx} \text{ at }x = \frac{\pi}{4}\] ?
Find \[\frac{dy}{dx}\] \[y = x^{\sin x} + \left( \sin x \right)^x\] ?
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 \[y = e^{x^{e^x}} + x^{e^{e^x}} + e^{x^{x^e}}\], prove that \[\frac{dy}{dx} = e^{x^{e^x}} \cdot x^{e^x} \left\{ \frac{e^x}{x} + e^x \cdot \log x \right\}+ x^{e^{e^x}} \cdot e^{e^x} \left\{ \frac{1}{x} + e^x \cdot \log x \right\} + e^{x^{x^e}} x^{x^e} \cdot x^{e - 1} \left\{ x + e \log x \right\}\]
If \[\frac{dy}{dx}\] when \[x = a \cos \theta \text{ and } y = b \sin \theta\] ?
If \[x = a \left( \frac{1 + t^2}{1 - t^2} \right) \text { and y } = \frac{2t}{1 - t^2}, \text { find } \frac{dy}{dx}\] ?
Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2}, - \frac{1}{2 \sqrt{2}} \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - 1}{x + 1} \right)\] with respect to \[\sin^{- 1} \left( 3x - 4 x^3 \right), \text { if }- \frac{1}{2} < x < \frac{1}{2}\] ?
Differentiate \[\tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\] with respect to \[\sqrt{1 - x^2},\text {if} - 1 < x < 1\] ?
Differential coefficient of sec(tan−1 x) is ______.
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
If \[f\left( x \right) = \sqrt{x^2 - 10x + 25}\] then the derivative of f (x) in the interval [0, 7] is ____________ .
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \cdot \frac{dy}{dx} + y \cos^2 x = 0\] ?
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
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 x = a cos nt − b sin nt and \[\frac{d^2 x}{dt} = \lambda x\] then find the value of λ ?
If y = a cos (loge x) + b sin (loge x), then x2 y2 + xy1 =
Find the height of a cylinder, which is open at the top, having a given surface area, greatest volume, and radius r.
