हिंदी

If Y = Cos − 1 ( 2 X ) + 2 Cos − 1 √ 1 − 4 X 2 , 0 < X < 1 2 , Find D Y D X . ? - Mathematics

Advertisements
Advertisements

प्रश्न

If  \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?

योग
Advertisements

उत्तर

\[\text{ Here, y }= \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}\]

\[\text{ Put 2x  }= \cos\theta\]

\[ \therefore y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \sqrt{1 - \cos^2 \theta}\]

\[ \Rightarrow y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \left( \sin\theta \right)\]

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

\[\text{Here}, 0 < x < \frac{1}{2}\]

\[ \Rightarrow 0 < 2x < 1\]

\[ \Rightarrow 0 < \cos\theta < 1\]

\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]

\[\text{and}\]

\[ \Rightarrow 0 > - \theta > - \frac{\pi}{2}\]

\[ \Rightarrow \frac{\pi}{2} > \left( \frac{\pi}{2} - \theta \right) > 0\]

\[ \Rightarrow 0 < \left( \frac{\pi}{2} - \theta \right) < \frac{\pi}{2}\]

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

\[ y = \theta + 2\left( \frac{\pi}{2} - \theta \right) .......\left[ Since, \cos^{- 1} \left( \cos\left( \theta \right) \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]

\[ \Rightarrow y = + \pi - 2\theta\]

\[ \Rightarrow y = \pi - \theta\]

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

Differentiate it with respect to x using chain rule,

\[\frac{d y}{d x} = 0 - \left[ \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right]\frac{d}{dx}\left( 2x \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]

\[ \therefore \frac{d y}{d x} = \frac{2}{\sqrt{1 - 4 x^2}}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.03 | Q 42 | पृष्ठ ६४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Differentiate sin (log x) ?


Differentiate \[\frac{e^x \log x}{x^2}\] ? 


Differentiate \[\left( \sin^{- 1} x^4 \right)^4\] ?


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


If \[y = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] .prove that \[\frac{dy}{dx} = 1 - y^2\] ?


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


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


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


Differentiate \[\sin^{- 1} \left( 2 x^2 - 1 \right), 0 < x < 1\]  ?


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


Differentiate 

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


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 \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] ?


If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?


If  \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find}  \frac{dy}{dx}\] ?


find  \[\frac{dy}{dx}\]  \[y = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right) \left( 4x - 1 \right)}}\] ?

 


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


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 `y = x^tan x + sqrt(x^2 + 1)/2, "find"  (dy)/(dx) ?`

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

 


If \[y = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}}\], prove that \[\frac{dy}{dx} = 2\ at\ x = \frac{\pi}{4}\] ?

 


If  \[x = 2 \cos \theta - \cos 2 \theta \text{ and y} = 2 \sin \theta - \sin 2 \theta\], prove that \[\frac{dy}{dx} = \tan \left( \frac{3 \theta}{2} \right)\] ?


Write the derivative of sinx with respect to cos x ?


Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2 \sqrt{2}}, \frac{1}{\sqrt{2 \sqrt{2}}} \right)\] ?

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


If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\] _________ .


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


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


If y = x + tan x, show that  \[\cos^2 x\frac{d^2 y}{d x^2} - 2y + 2x = 0\] ?


If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0 ?


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


If x = 4z2 + 5, y = 6z2 + 7z + 3, find \[\frac{d^2 y}{d x^2}\] ?


If y = 3 e2x + 2 e3x, prove that  \[\frac{d^2 y}{d x^2} - 5\frac{dy}{dx} + 6y = 0\] ?


\[\text { If x } = a\left( \cos t + t \sin t \right) \text { and y} = a\left( \sin t - t \cos t \right),\text { then find the value of } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?


\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]

 


If \[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\] then (1 − x)2 '' (x) − xf(x) =

 


If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?


If y = xn−1 log x then x2 y2 + (3 − 2n) xy1 is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×