मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज

उत्तर

\[\text{ We have, }x^y + y^x = \left( x + y \right)^{x + y} \]

\[ \Rightarrow e^{ \log x^y} + e^{\log y^x } = e^{ \log \left( x + y \right)^\left( x + y \right) } \]

\[ \Rightarrow e^{y \log x} + e^{x \log y} = e^{ \left( x + y \right) \log\left( x + y \right) }\]

Differentiating with respect to x using chain rule and product rule,

\[\Rightarrow \frac{d}{dx}\left( e^{y \log x} \right) + \frac{d}{dx}\left( e^{x \log y} \right) = \frac{d}{dx} e^{\left( x + y \right)\log\left( x + y \right)} \]

\[ \Rightarrow e^{y \log x } \left[ y\frac{d}{dx}\left( \log x \right) + \log x\frac{dy}{dx} \right] + e^{x \log y} \left[ x\frac{d}{dx}\log y + \log y\frac{d}{dx}\left( x \right) \right] = e^\left( x + y \right)\log\left( x + y \right) \frac{d}{dx}\left[ \left( x + y \right)\log\left( x + y \right) \right]\]

\[ \Rightarrow e^{ \log x^y } \left[ y\left( \frac{1}{x} \right) + \log x\frac{dy}{dx} \right] + e^{ \log x } \left[ \frac{x}{y}\frac{dy}{dx} + \log y\left( 1 \right) \right] = e^{{\log }\left( x + y \right)^\left( x + y \right)} \left[ \left( x + y \right)\frac{d}{dx}\log\left( x + y \right) + \log\left( x + y \right)\frac{d}{dx}\left( x + y \right) \right]\]

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

\[ \Rightarrow x^y \times \frac{y}{x} + x^y \log x\frac{dy}{dx} + y^x \times \frac{x}{y}\frac{dy}{dx} + y^x \log y = \left( x + y \right)^\left( x + y \right) \left[ 1 \times \left( 1 + \frac{dy}{dx} \right) + \log\left( x + y \right)\left( 1 + \frac{dy}{dx} \right) \right]\]

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

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

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.05 [पृष्ठ ८९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.05 | Q 38 | पृष्ठ ८९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


Differentiate the following functions from first principles log cos x ?


Differentiate etan x ?


Differentiate (log sin x)?


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


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


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


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


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


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


If \[xy = 1\] prove that \[\frac{dy}{dx} + y^2 = 0\] ?


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


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


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


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


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

 


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

 


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

 


Find \[\frac{dy}{dx}\] , when \[x = b   \sin^2   \theta  \text{ and }  y = a   \cos^2   \theta\] ?


Find \[\frac{dy}{dx}\] , when  \[x = \cos^{- 1} \frac{1}{\sqrt{1 + t^2}} \text{ and y } = \sin^{- 1} \frac{t}{\sqrt{1 + t^2}}, t \in R\] ?


Find  \[\frac{dy}{dx}\] , when  \[x = \frac{1 - t^2}{1 + t^2} \text{ and y } = \frac{2 t}{1 + t^2}\] ?

 


If \[x = \sin^{- 1} \left( \frac{2 t}{1 + t^2} \right) \text{ and y } = \tan^{- 1} \left( \frac{2 t}{1 - t^2} \right), - 1 < t < 1\] porve that \[\frac{dy}{dx} = 1\] ?

 


\[\sin x = \frac{2t}{1 + t^2}, \tan y = \frac{2t}{1 - t^2}, \text { find }  \frac{dy}{dx}\] ?

If \[x = 3\sin t - \sin3t, y = 3\cos t - \cos3t \text{ find }\frac{dy}{dx} \text{ at } t = \frac{\pi}{3}\] ?


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


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


If \[y = \sqrt{\sin x + y},\text { then } \frac{dy}{dx} =\] __________ .


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 x cos x ?


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


If \[y = e^{2x} \left( ax + b \right)\]  show that  \[y_2 - 4 y_1 + 4y = 0\] ?


\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text {  and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?


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


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


If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

 


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

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×