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

Prove that `y=(4sintheta)/(2+costheta)-theta `


If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60º.


Differentiate log7 (2x − 3) ?


Differentiate (log sin x)?


Differentiate \[e^\sqrt{\cot x}\] ?


Differentiate  \[e^x \log \sin 2x\] ?


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


Differentiate \[\cos^{- 1} \left\{ \frac{\cos x + \sin x}{\sqrt{2}} \right\}, - \frac{\pi}{4} < x < \frac{\pi}{4}\] ?


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


Differentiate \[\tan^{- 1} \left( \frac{5 x}{1 - 6 x^2} \right), - \frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}\] ?


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 \[xy = c^2\]  ?


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 \[\sin^2 y + \cos xy = k,\] find  \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\] 


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


Differentiate \[{10}^\left( {10}^x \right)\] ?


Differentiate  \[\sin \left( x^x \right)\] ?


Differentiate \[\left( \tan x \right)^{1/x}\] ?


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

 


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


If \[x^{16} y^9 = \left( x^2 + y \right)^{17}\] ,prove that \[x\frac{dy}{dx} = 2 y\] ?


If \[e^x + e^y = e^{x + y}\] , prove that

\[\frac{dy}{dx} + e^{y - x} = 0\] ?


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


If `y = x^tan x + sqrt(x^2 + 1)/2, "find"  (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  \[x = 2 \cos \theta - \cos 2 \theta \text{ and y} = 2 \sin \theta - \sin 2 \theta\], prove that \[\frac{dy}{dx} = \tan \left( \frac{3 \theta}{2} \right)\] ?


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

 


If \[f\left( x \right) = x + 1\] , then write the value of \[\frac{d}{dx} \left( fof \right) \left( x \right)\] ?


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


If \[y = \left( 1 + \frac{1}{x} \right)^x , \text{then} \frac{dy}{dx} =\] ____________.


For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text {  at } \left( 1/4, 1/4 \right)\text {  is }\] _____________ .


If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .


The derivative of \[\cos^{- 1} \left( 2 x^2 - 1 \right)\] with respect to  \[\cos^{- 1} x\]  is ___________ .


Find the second order derivatives of the following function  x3 + tan x ?


Find the second order derivatives of the following function e6x cos 3x  ?


If y = x3 log x, prove that \[\frac{d^4 y}{d x^4} = \frac{6}{x}\] ?


If \[y = \frac{\log x}{x}\] show that \[\frac{d^2 y}{d x^2} = \frac{2 \log x - 3}{x^3}\] ?


If \[y = e^{\tan^{- 1} x}\] prove that (1 + x2)y2 + (2x − 1)y1 = 0 ?


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×