हिंदी

If X 13 Y 7 = ( X + Y ) 20 Prove that D Y D X = Y X ? - Mathematics

Advertisements
Advertisements

प्रश्न

If \[x^{13} y^7 = \left( x + y \right)^{20}\] prove that \[\frac{dy}{dx} = \frac{y}{x}\] ?

Advertisements

उत्तर

\[\text{ We have,} x^{13} y^7 = \left( x + y \right)^{20} \]

Taking log on both sides,

\[\log\left( x^{13} y^7 \right) = \log \left( x + y \right)^{20} \]

\[ \Rightarrow 13\log x + 7\log y = 20\log\left( x + y \right)\]

Differentiating with respect to x using chain rule,

\[13\frac{d}{dx}\left( \log x \right) + 7\frac{d}{dx}\left( \log y \right) = 20\frac{d}{dx}\log\left( x + y \right)\]

\[ \Rightarrow \frac{13}{x} + \frac{7}{y}\frac{dy}{dx} = \frac{20}{x + y}\frac{d}{dx}\left( x + y \right)\]

\[ \Rightarrow \frac{13}{x} + \frac{7}{y}\frac{dy}{dx} = \frac{20}{x + y}\left[ 1 + \frac{dy}{dx} \right]\]

\[ \Rightarrow \frac{7}{y}\frac{dy}{dx} - \frac{20}{x + y}\frac{dy}{dx} = \frac{20}{x + y} - \frac{13}{x}\]

\[ \Rightarrow \frac{dy}{dx}\left[ \frac{7}{y} - \frac{20}{x + y} \right] = \frac{20}{x + y} - \frac{13}{x}\]

\[ \Rightarrow \frac{dy}{dx}\left[ \frac{7\left( x + y \right) - 20y}{y\left( x + y \right)} \right] = \left[ \frac{20x - 13\left( x + y \right)}{x\left( x + y \right)} \right]\]

\[ \Rightarrow \frac{dy}{dx} = \left[ \frac{20x - 13x - 13y}{x\left( x + y \right)} \right]\left[ \frac{y\left( x + y \right)}{7x + 7y - 20y} \right]\]

\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x}\left( \frac{7x - 13y}{7x - 13y} \right)\]

\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.05 [पृष्ठ ८९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.05 | Q 33 | पृष्ठ ८९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Differentiate the following functions from first principles e3x.


​Differentiate the following function from first principles \[e^\sqrt{\cot x}\] .


Differentiate tan2 x ?


Differentiate \[3^{x^2 + 2x}\] ?


If \[y = e^x \cos x\] ,prove that \[\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)\] ?


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


Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?


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


If \[y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0\] ,prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ? 


Find  \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?

 


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


Find  \[\frac{dy}{dx}\] \[y = e^x + {10}^x + x^x\] ?

 


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


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


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


\[\text{If y} = 1 + \frac{\alpha}{\left( \frac{1}{x} - \alpha \right)} + \frac{{\beta}/{x}}{\left( \frac{1}{x} - \alpha \right)\left( \frac{1}{x} - \beta \right)} + \frac{{\gamma}/{x^2}}{\left( \frac{1}{x} - \alpha \right)\left( \frac{1}{x} - \beta \right)\left( \frac{1}{x} - \gamma \right)}, \text{ find } \frac{dy}{dx}\] is:

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

 


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


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

 


Differentiate x2 with respect to x3


Differentiate\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\] with respect to \[\sin^{-1} \left( \frac{2x}{1 + x^2} \right)\], If \[- 1 < x < 1, x \neq 0 .\] ?


If f (x) = loge (loge x), then write the value of `f' (e)` ?


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


If \[f'\left( x \right) = \sqrt{2 x^2 - 1} \text { and y } = f \left( x^2 \right)\] then find \[\frac{dy}{dx} \text { at } x = 1\] ?


If \[f\left( x \right) = \log \left\{ \frac{u \left( x \right)}{v \left( x \right)} \right\}, u \left( 1 \right) = v \left( 1 \right) \text{ and }u' \left( 1 \right) = v' \left( 1 \right) = 2\] , then find the value of `f' (1)` ?


Differential coefficient of sec(tan−1 x) is ______.


Given  \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .


Let  \[\cup = \sin^{- 1} \left( \frac{2 x}{1 + x^2} \right) \text { and }V = \tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text { then } \frac{d \cup}{dV} =\] ____________ .


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


If y = ex cos x, prove that \[\frac{d^2 y}{d x^2} = 2 e^x \cos \left( x + \frac{\pi}{2} \right)\] ?


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 \[y = \left[ \log \left( x + \sqrt{x^2 + 1} \right) \right]^2\] show that \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\] ?


If y = (tan−1 x)2, then prove that (1 + x2)2 y2 + 2x(1 + x2)y1 = 2 ?


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


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


If \[y = \tan^{- 1} \left\{ \frac{\log_e \left( e/ x^2 \right)}{\log_e \left( e x^2 \right)} \right\} + \tan^{- 1} \left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)\], then \[\frac{d^2 y}{d x^2} =\]

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×