मराठी

If Sin ( X Y ) + Y X = X 2 − Y 2 , Find D Y D X ? - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[\text{ We have, }\sin\left( xy \right) + \frac{y}{x} = x^2 - y^2\]

Differentiating with respect to x, we get,

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

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

\[ \Rightarrow \cos\left( xy \right)\left\{ x\frac{dy}{dx} + y\frac{d}{dx}\left( x \right) \right\} + \left\{ \frac{x\frac{dy}{dx} - y\left( 1 \right)}{x^2} \right\} = 2x - 2y\frac{dy}{dx}\]

\[ \Rightarrow \cos\left( xy \right)\left\{ x\frac{dy}{dx} + y\left( 1 \right) \right\} + \frac{1}{x^2}\left( x\frac{dy}{dx} - y \right) = 2x - 2y\frac{dy}{dx}\]

\[ \Rightarrow x \cos\left( xy \right)\frac{dy}{dx} + y \cos\left( xy \right) + \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = 2x - 2y\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx}\left\{ x \cos\left( xy \right) + \frac{1}{x} + 2y \right\} = \frac{y}{x^2} - y \cos\left( xy \right) + 2x\]

\[ \Rightarrow \frac{dy}{dx}\left\{ \frac{x^2 \cos\left( xy \right) + 1 + 2xy}{x} \right\} = \frac{1}{x^2}\left( y - x^2 y \cos\left( xy \right) + 2 x^3 \right)\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2 x^3 + y - x^2 y \cos\left( xy \right)}{x\left( x^2 \cos\left( xy \right) + 1 + 2xy \right)}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.04 [पृष्ठ ७५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.04 | Q 25 | पृष्ठ ७५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If the function f(x)=2x39mx2+12m2x+1, where m>0 attains its maximum and minimum at p and q respectively such that p2=q, then find the value of m.

 


Differentiate \[e^{\tan 3 x} \] ?


Differentiate \[\log \left( \cos x^2 \right)\] ?


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


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


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


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

 


If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?


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


Differentiate \[x^{\sin x}\]  ?


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


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


Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?


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

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


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


Find \[\frac{dy}{dx}\] ,When \[x = a \left( 1 - \cos \theta \right) \text{ and } y = a \left( \theta + \sin \theta \right) \text{ at } \theta  = \frac{\pi}{2}\] ?


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

 


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


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


The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]


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


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


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


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


If x = a sec θ, y = b tan θ, prove that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?


If x = a (1 − cos3θ), y = a sin3θ, prove that \[\frac{d^2 y}{d x^2} = \frac{32}{27a} \text { at } \theta = \frac{\pi}{6}\]?


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


Find \[\frac{d^2 y}{d x^2}\] where \[y = \log \left( \frac{x^2}{e^2} \right)\] ?


If y = cosec−1 xx >1, then show that \[x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\] ?


\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] Disclaimer: There is a misprint in the question. It must be 

\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] instead of 1

\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] ?


If x = a cos nt − b sin nt and \[\frac{d^2 x}{dt} = \lambda x\]  then find the value of λ ?


If x = t2 and y = t3, find \[\frac{d^2 y}{d x^2}\] ?


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


If xy − loge y = 1 satisfies the equation \[x\left( y y_2 + y_1^2 \right) - y_2 + \lambda y y_1 = 0\]

 


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


Differentiate sin(log sin x) ?


Differentiate the following with respect to x

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×