English

If Xy = E(X – Y), Then Show that D Y D X = Y ( X − 1 ) X ( Y + 1 ) . - Mathematics

Advertisements
Advertisements

Question

If xy = e(x – y), then show that `dy/dx = (y(x-1))/(x(y+1)) .`

Advertisements

Solution

We have , `xy = e^(x-y)`

Taking log on both sides,

`log (xy) = log (e^(x-y))`

⇒ `log x + log y= (x-y)log e`

⇒ `logx + log y = (x - y) xx 1`

⇒ `log x+ log y = x-y`

⇒ `d/dx (log x) + d/dx (log y ) = d/(dr) (x) - dy/dx`

⇒ `1/x + 1/y dy/dx = 1 -dx /dx`

⇒` ( 1 +1/y) dy/dx = 1 -1/x`

⇒ `(y+1/y)dy/dx = (x-1)/x`

⇒ `dy/dx = (y(x-1))/(x(y+1))`

Hence proved.

shaalaa.com
  Is there an error in this question or solution?
2016-2017 (March) Foreign Set 3

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Differentiate the following functions from first principles  \[e^\sqrt{2x}\].


Differentiate \[\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}\] ?


Differentiate \[\cos^{- 1} \left\{ \frac{\cos x + \sin x}{\sqrt{2}} \right\}, - \frac{\pi}{4} < x < \frac{\pi}{4}\] ?


Differentiate \[\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0\] ?


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

 


If the derivative of tan−1 (a + bx) takes the value 1 at x = 0, prove that 1 + a2 = b ?


Find  \[\frac{dy}{dx}\] in the following case \[x^5 + y^5 = 5 xy\] ?

 


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


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


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


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


Differentiate \[x^{\sin^{- 1} x}\]  ?


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


\[\text{ If }\cos y = x\cos\left( a + y \right),\text{  where } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?

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

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


\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?


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 \[y = \sec^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right)\] then write the value of \[\frac{dy}{dx} \] ?


If \[y = \log \left| 3x \right|, x \neq 0, \text{ find } \frac{dy}{dx} \] ? 


If \[f\left( x \right) = \left| x^2 - 9x + 20 \right|\]  then `f' (x)` is equal to ____________ .


If \[f\left( x \right) = \left( \frac{x^l}{x^m} \right)^{l + m} \left( \frac{x^m}{x^n} \right)^{m + n} \left( \frac{x^n}{x^l} \right)^{n + 1}\] the f' (x) is equal to _____________ .


If \[y = \log \sqrt{\tan x}\] then the value of \[\frac{dy}{dx}\text { at }x = \frac{\pi}{4}\] is given by __________ .


If y = ex cos x, show that \[\frac{d^2 y}{d x^2} = 2 e^{- x} \sin x\] ?


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


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 `log [x+2+sqrt(x^2+4x+1)]`


Range of 'a' for which x3 – 12x + [a] = 0 has exactly one real root is (–∞, p) ∪ [q, ∞), then ||p| – |q|| is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×