मराठी

Differentiate Tan − 1 { X a + √ a 2 − X 2 } , − a < X < a ? - Mathematics

Advertisements
Advertisements

प्रश्न

Differentiate \[\tan^{- 1} \left\{ \frac{x}{a + \sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?

बेरीज
Advertisements

उत्तर

\[\text{ Let, y } = \tan^{- 1} \left\{ \frac{x}{a + \sqrt{a^2 - x^2}} \right\}\]

\[\text{ Put x }= a \sin\theta\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a + \sqrt{a^2 - a^2 \sin^2 \theta}} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left( \frac{a \sin\theta}{a + \sqrt{a^2 \left( 1 - \sin^2 \theta \right)}} \right) \]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a + a \cos\theta} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a\left( 1 + \cos\theta \right)} \right\} \]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{\sin\theta}{1 + \cos\theta} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left( \tan \frac{\theta}{2} \right) . . . \left( i \right) \]

\[\text{Here }, - a < x < a\]

\[ \Rightarrow - 1 < \frac{x}{a} < 1\]

\[ \Rightarrow - 1 < \sin\theta < 1\]

\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2}\]

\[ \Rightarrow - \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\]

\[\text{ So, from equation } \left( i \right), \]

\[ y = \frac{\theta}{2} .......\left[ \text{ Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]

\[ \Rightarrow y = \frac{1}{2} \sin^{- 1} \left( \frac{x}{a} \right) ..........\left[ \text{ Since }, x = a \sin\theta \right]\]

\[\text{ Differentiating it with respect to x }, \]

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

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

\[ \therefore \frac{d y}{d x} = \frac{1}{2\sqrt{a^2 - x^2}}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.03 | Q 13 | पृष्ठ ६३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Differentiate the following functions from first principles e−x.


\[\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\] ?


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


Differentiate \[\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0\] ?


Differentiate \[\tan^{- 1} \left( \frac{x}{1 + 6 x^2} \right)\] ?


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 \[x y^2 = 1,\] prove that \[2\frac{dy}{dx} + y^3 = 0\] ?


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 \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?


If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find}  \frac{dy}{dx}\] ?


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


Find \[\frac{dy}{dx}\]  \[y = x^n + n^x + x^x + n^n\] ?

Find  \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?

 


Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\log x} + \cos^2 \left( \frac{\pi}{4} \right)\] ?


If \[y = \sin \left( x^x \right)\] prove that  \[\frac{dy}{dx} = \cos \left( x^x \right) \cdot x^x \left( 1 + \log x \right)\] ?


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 \[y^x + x^y + x^x = a^b\] ,find \[\frac{dy}{dx}\] ?


If \[y = \sqrt{\cos x + \sqrt{\cos x + \sqrt{\cos x + . . . to \infty}}}\] , prove that \[\frac{dy}{dx} = \frac{\sin x}{1 - 2 y}\] ?


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


Differentiate \[\cos^{- 1} \left( 4 x^3 - 3x \right)\] with respect to \[\tan^{- 1} \left( \frac{\sqrt{1 - x^2}}{x} \right), \text{ if }\frac{1}{2} < x < 1\] ? 


If \[f'\left( 1 \right) = 2 \text { and y } = f \left( \log_e x \right), \text { find} \frac{dy}{dx} \text { at }x = e\] ?


If \[f\left( 1 \right) = 4, f'\left( 1 \right) = 2\] find the value of the derivative of  \[\log \left( f\left( e^x \right) \right)\] w.r. to x at the point x = 0 ?

 


If \[\left| x \right| < 1 \text{ and y} = 1 + x + x^2 + . . \]  to ∞, then find the value of  \[\frac{dy}{dx}\] ?


\[\frac{d}{dx} \left\{ \tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right) \right\} \text { equals }\] ______________ .


\[\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\} \right]\] equals ___________ .

If \[y = \sqrt{\sin x + y},\text { then } \frac{dy}{dx} =\] __________ .


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 x3 log ?


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 y = ex (sin + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?


\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?


\[\text { If }y = A e^{- kt} \cos\left( pt + c \right), \text { prove that } \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 \] ?


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 

 


\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]

\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]


Differentiate the following with respect to x

\[\cot^{- 1} \left( \frac{1 - x}{1 + x} \right)\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×