हिंदी

If X Sin ( a + Y ) + Sin a Cos ( a + Y ) = 0 Prove that D Y D X = Sin 2 ( a + Y ) Sin a ? - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[\text{ We have, } x\sin\left( a + y \right) + \sin a\cos\left( a + y \right) = 0 \]

Differentiate with respect to x, 

\[\Rightarrow \frac{d}{dx}\left[ x \sin\left( a + y \right) \right] + \frac{d}{dx}\left[ \sin a \cos\left( a + y \right) \right] = 0\]

\[ \Rightarrow \left[ x\frac{d}{dx}\sin \left( a + y \right) + \sin\left( a + y \right)\frac{d}{dx}\left( x \right) \right] + \sin a\frac{d}{dx}\cos\left( a + y \right) = 0 \]

\[ \Rightarrow \left[ x \cos\left( a + y \right)\frac{d}{dx}\left( a + y \right) + \sin\left( a + y \right)\left( 1 \right) \right] + \sin a\left[ - \sin\left( a + y \right)\frac{d}{dx}\left( a + y \right) \right] = 0\]

\[ \Rightarrow x \cos\left( a + y \right)\frac{d y}{d x} + \sin\left( a + y \right) - \sin a\sin\left( a + y \right)\frac{d y}{d x} = 0\]

\[ \Rightarrow \frac{d y}{d x}\left[ x \cos\left( a + y \right) - \sin a \sin\left( a + y \right) \right] = - \sin\left( a + y \right)\]

\[ \Rightarrow \frac{d y}{d x}\left[ - \sin a\frac{\cos^2 \left( a + y \right)}{\sin\left( a + y \right)} - \sin a \sin\left( a + y \right) \right] = - \sin\left( a + y \right) \left[ \because x = - \sin a\frac{\cos\left( a + y \right)}{\sin\left( a + y \right)} \right]\]

\[ \Rightarrow - \frac{d y}{d x}\left[ \frac{\sin a \cos^2 \left( a + y \right) + \sin a \sin^2 \left( a + y \right)}{\sin\left( a + y \right)} \right] = - \sin\left( a + y \right)\]

\[ \Rightarrow \frac{d y}{d x} = \sin\left( a + y \right)\left[ \frac{\sin\left( a + y \right)}{\sin a\left\{ \cos^2 \left( a + y \right) + \sin^2 \left( a + y \right) \right\}} \right]\]

\[ \Rightarrow \frac{d y}{d x} = \frac{\sin^2 \left( a + y \right)}{\sin a} \]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.04 [पृष्ठ ७५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.04 | Q 22 | पृष्ठ ७५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Differentiate the following functions from first principles x2ex ?


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


Differentiate  \[e^x \log \sin 2x\] ?


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


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


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


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


Differentiate \[\sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x \in R\] ?


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


Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.


If \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\] \[\frac{dy}{dx} \] ?


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


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


Differentiate \[\left( 1 + \cos x \right)^x\] ?


Differentiate \[e^{x \log x}\] ?


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


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


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


If \[\left( \cos x \right)^y = \left( \tan y \right)^x\] , prove that \[\frac{dy}{dx} = \frac{\log \tan y + y \tan x}{ \log \cos x - x \sec y \ cosec\ y }\] ?


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

 


If \[x = 10 \left( t - \sin t \right), y = 12 \left( 1 - \cos t \right), \text { find } \frac{dy}{dx} .\] ?

 


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


Differentiate \[\tan^{- 1} \left( \frac{1 + ax}{1 - ax} \right)\] with respect to \[\sqrt{1 + a^2 x^2}\] ?


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


If f (x) is an even function, then write whether `f' (x)` is even or odd ?


If \[x = a \cos^3 \theta, y = a \sin^3 \theta, \text { then } \sqrt{1 + \left( \frac{dy}{dx} \right)^2} =\] ____________ .


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


If  \[\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3 \left( x^3 - y^3 \right)\] then \[\frac{dy}{dx}\] is equal to ____________ .


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


If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \cdot \frac{dy}{dx} + y \cos^2 x = 0\] ?


If y = (sin−1 x)2, prove that (1 − x2)

\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?


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 ?


If y = cot x show that \[\frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\] ?


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


If y = etan x, then (cos2 x)y2 =


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×