हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Differentiate `2^(x^3)` ?


Differentiate \[\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}\] ?


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


If \[y = \sqrt{x^2 + a^2}\] prove that  \[y\frac{dy}{dx} - x = 0\] ?


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


Differentiate \[\sin^{- 1} \left\{ \frac{\sqrt{1 + x} + \sqrt{1 - x}}{2} \right\}, 0 < x < 1\] ?


Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] with respect to x.


Find  \[\frac{dy}{dx}\] in the following case \[e^{x - y} = \log \left( \frac{x}{y} \right)\] ?

 


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


If \[\sqrt{y + x} + \sqrt{y - x} = c, \text {show that } \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\] ?


Differentiate \[x^{\sin x}\]  ?


Find \[\frac{dy}{dx}\], When \[x = a \left( \theta + \sin \theta \right) \text{ and } y = a \left( 1 - \cos \theta \right)\] ?


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


If \[x = \cos t \text{ and y }  = \sin t,\] prove that  \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?

 


If  \[x = a\left( t + \frac{1}{t} \right) \text{ and y } = a\left( t - \frac{1}{t} \right)\] ,prove that  \[\frac{dy}{dx} = \frac{x}{y}\]?

 


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


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


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 \[y = \sin^{- 1} x + \cos^{- 1} x\] ,find \[\frac{dy}{dx}\] ?


If \[y = x^x , \text{ find } \frac{dy}{dx} \text{ at } x = e\] ?


Differential coefficient of sec(tan−1 x) is ______.


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 \[f\left( x \right) = \sqrt{x^2 - 10x + 25}\]  then the derivative of f (x) in the interval [0, 7] is ____________ .


If \[f\left( x \right) = \left( \frac{x^l}{x^m} \right)^{l + m} \left( \frac{x^m}{x^n} \right)^{m + n} \left( \frac{x^n}{x^l} \right)^{n + 1}\] the f' (x) is equal to _____________ .


Find the second order derivatives of the following function  log (log x)  ?


If x = a (θ + sin θ), y = a (1 + cos θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{a}{y^2}\] ?


If x = sin ty = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?


If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0 ?


If y = (tan−1 x)2, then prove that (1 + x2)2 y2 + 2x(1 + x2)y1 = 2 ?


\[\text { Find A and B so that y = A } \sin3x + B \cos3x \text { satisfies the equation }\]

\[\frac{d^2 y}{d x^2} + 4\frac{d y}{d x} + 3y = 10 \cos3x \] ?


If x = f(t) and y = g(t), then write the value of \[\frac{d^2 y}{d x^2}\] ?


If y = a sin mx + b cos mx, then \[\frac{d^2 y}{d x^2}\]   is equal to

 


If x = f(t) and y = g(t), then \[\frac{d^2 y}{d x^2}\] is equal to

 


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


f(x) = 3x2 + 6x + 8, x ∈ R


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×