मराठी

If Y √ X 2 + 1 = Log ( √ X 2 + 1 − X ) ,Show that ( X 2 + 1 ) D Y D X + X Y + 1 = 0 ? - Mathematics

Advertisements
Advertisements

प्रश्न

If \[y \sqrt{x^2 + 1} = \log \left( \sqrt{x^2 + 1} - x \right)\] ,Show that \[\left( x^2 + 1 \right) \frac{dy}{dx} + xy + 1 = 0\] ?

Advertisements

उत्तर

\[\text{ We have, y} \sqrt{x^2 + 1} = \log\left( \sqrt{x^2 + 1} - x \right)\] 

Differentiating with respect to x, we get,

\[\Rightarrow \frac{d}{dx}\left( y\sqrt{x^2 + 1} \right) = \frac{d}{dx}\log\left( \sqrt{x^2 + 1} - x \right) \left[ \text{ using product rule and chain rule } \right]\]

\[ \Rightarrow y\frac{d}{dx}\left( \sqrt{x^2 + 1} \right) + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)} \times \frac{d}{dx}\left( \sqrt{x^2 + 1} - x \right)\]

\[ \Rightarrow \frac{y}{2\sqrt{x^2 + 1}} \times \frac{d}{dx}\left( x^2 + 1 \right) + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)} \times \left[ \frac{1}{2\sqrt{x^2 + 1}}\frac{d}{dx}\left( x^2 + 1 \right) - 1 \right]\]

\[ \Rightarrow \frac{2xy}{2\sqrt{x^2 + 1}} + \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{1}{\left( \sqrt{x^2 + 1} - x \right)}\left[ \frac{2x}{2\sqrt{x^2 + 1}} - 1 \right]\]

\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \left[ \frac{1}{\sqrt{x^2 + 1} - x} \right]\left[ \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right] - \frac{xy}{\sqrt{x^2 + 1}}\]

\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{- 1}{\sqrt{x^2 + 1}} - \frac{xy}{\sqrt{x^2 + 1}}\]

\[ \Rightarrow \sqrt{x^2 + 1}\frac{dy}{dx} = \frac{- \left( 1 + xy \right)}{\sqrt{x^2 + 1}}\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{dy}{dx} = - \left( 1 + xy \right)\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{dy}{dx} + 1 + xy = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.04 [पृष्ठ ७५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.04 | Q 24 | पृष्ठ ७५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Differentiate tan 5x° ?


Differentiate (log sin x)?


Differentiate \[\sqrt{\frac{1 + x}{1 - x}}\] ?


Differentiate \[\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}\] ?


Differentiate \[\sin \left( 2 \sin^{- 1} x \right)\] ?


If \[y = \frac{x}{x + 2}\]  , prove tha \[x\frac{dy}{dx} = \left( 1 - y \right) y\] ? 


Differentiate \[\cos^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?


Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?


Differentiate 

\[\tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right), \frac{\pi}{4} < x < \frac{\pi}{4}\] ?


If \[y = \sin^{- 1} \left( \frac{x}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right), 0 < x < \infty\] prove that  \[\frac{dy}{dx} = \frac{2}{1 + x^2} \] ?

 


If  \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?

 


Find  \[\frac{dy}{dx}\] in the following case \[\left( x + y \right)^2 = 2axy\] ?

 


If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\]  ?


If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{y}{x}\] ?


If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?


If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?


Differentiate \[x^{1/x}\]  with respect to x.


Differentiate \[\left( \sin^{- 1} x \right)^x\] ?


If \[y^x = e^{y - x}\] ,prove that \[\frac{dy}{dx} = \frac{\left( 1 + \log y \right)^2}{\log y}\] ?


If \[e^{x + y} - x = 0\] ,prove that \[\frac{dy}{dx} = \frac{1 - x}{x}\] ?


Find \[\frac{dy}{dx}\],when \[x = a e^\theta \left( \sin \theta - \cos \theta \right), y = a e^\theta \left( \sin \theta + \cos \theta \right)\] ?


Find \[\frac{dy}{dx}\] when \[x = \frac{2 t}{1 + t^2} \text{ and } y = \frac{1 - t^2}{1 + t^2}\] ?


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 \[x = \sin^{- 1} \left( \frac{2 t}{1 + t^2} \right) \text{ and y } = \tan^{- 1} \left( \frac{2 t}{1 - t^2} \right), - 1 < t < 1\] porve that \[\frac{dy}{dx} = 1\] ?

 


Differentiate (log x)x with respect to log x ?


Let g (x) be the inverse of an invertible function f (x) which is derivable at x = 3. If f (3) = 9 and `f' (3) = 9`, write the value of `g' (9)`.


If \[\pi \leq x \leq 2\pi \text { and y } = \cos^{- 1} \left( \cos x \right), \text { find } \frac{dy}{dx}\] ?


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 \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .


If \[f\left( x \right) = \sqrt{x^2 + 6x + 9}, \text { then } f'\left( x \right)\] is equal to ______________ .


If \[f\left( x \right) = \left| x^2 - 9x + 20 \right|\]  then `f' (x)` is equal to ____________ .


If \[f\left( x \right) = \left| x - 3 \right| \text { and }g\left( x \right) = fof \left( x \right)\]  is equal to __________ .


If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then  } \frac{dy}{dx}\] is equal to ___________ .


Find the second order derivatives of the following function ex sin 5x  ?


If y = x + ex, find \[\frac{d^2 x}{d y^2}\] ?


If x = t2, y = t3, then \[\frac{d^2 y}{d x^2} =\] 

 


If \[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x\] then the value of ar, 0 < r ≤ n, is equal to 

 


If xy = e(x – y), then show that `dy/dx = (y(x-1))/(x(y+1)) .`


If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]


Differentiate `log [x+2+sqrt(x^2+4x+1)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×