Advertisements
Advertisements
Question
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
Advertisements
Solution
\[\text{We have, xy } = 4\]
\[ \Rightarrow y = \frac{4}{x}\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{4}{x} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\frac{d}{dx}\left( x^{- 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\left( - 1 \times x^{- 1 - 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\left( - \frac{1}{x^2} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- 4}{x^2}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{4}{\left( \frac{4}{y} \right)^2} \left[ \because x = \frac{4}{y} \right]\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{4 y^2}{16}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{y^2}{4}\]
\[ \Rightarrow 4\frac{d y}{d x} = - y^2 \]
\[ \Rightarrow 4\frac{d y}{d x} = 3 y^2 - 4 y^2 \]
\[ \Rightarrow 4\frac{d y}{d x} + 4 y^2 = 3 y^2 \]
\[ \Rightarrow 4\left( \frac{d y}{d x} + y^2 \right) = 3 y^2 \]
Dividing both side by x,
\[\Rightarrow \frac{4}{x}\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}\]
\[ \Rightarrow y\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x} \]
\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{y}\]
\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = 3y\]
APPEARS IN
RELATED QUESTIONS
Differentiate the following functions from first principles e3x.
Differentiate \[e^{\sin^{- 1} 2x}\] ?
\[\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\] ?
If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\], prove that \[2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}\] ?
Differentiate \[\sin^{- 1} \left( 2 x^2 - 1 \right), 0 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( 1 - 2 x^2 \right), 0 < x < 1\] ?
Differentiate \[\cos^{- 1} \left( \frac{1 - x^{2n}}{1 + x^{2n}} \right), < x < \infty\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\] ?
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 \[\sqrt{1 - x^2} + \sqrt{1 - y^2} = a \left( x - y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - x^2}\] ?
If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?
Differentiate \[e^{x \log x}\] ?
If \[x^y + y^x = \left( x + y \right)^{x + y} , \text{ find } \frac{dy}{dx}\] ?
If \[\left( \sin x \right)^y = x + y\] , prove that \[\frac{dy}{dx} = \frac{1 - \left( x + y \right) y \cot x}{\left( x + y \right) \log \sin x - 1}\] ?
Find \[\frac{dy}{dx}\],when \[x = a e^\theta \left( \sin \theta - \cos \theta \right), y = a e^\theta \left( \sin \theta + \cos \theta \right)\] ?
Find \[\frac{dy}{dx}\] ,when \[x = \frac{e^t + e^{- t}}{2} \text{ and } y = \frac{e^t - e^{- t}}{2}\] ?
Find \[\frac{dy}{dx}\] , when \[x = \frac{1 - t^2}{1 + t^2} \text{ and y } = \frac{2 t}{1 + t^2}\] ?
If \[x = \frac{1 + \log t}{t^2}, y = \frac{3 + 2\log t}{t}, \text { find } \frac{dy}{dx}\] ?
Differentiate x2 with respect to x3
Differentiate \[\tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\] with respect to \[\sqrt{1 - x^2},\text {if} - 1 < x < 1\] ?
If \[y = \log_a x, \text{ find } \frac{dy}{dx} \] ?
If \[y = \left( 1 + \frac{1}{x} \right)^x , \text{then} \frac{dy}{dx} =\] ____________.
If \[y = \sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] _____________ .
If \[f\left( x \right) = \left| x - 3 \right| \text { and }g\left( x \right) = fof \left( x \right)\] is equal to __________ .
If \[\sin y = x \cos \left( a + y \right), \text { then } \frac{dy}{dx}\] is equal to ______________ .
If y = x + tan x, show that \[\cos^2 x\frac{d^2 y}{d x^2} - 2y + 2x = 0\] ?
If x = a cos θ, y = b sin θ, show that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
\[\text { If x } = a\left( \cos t + t \sin t \right) \text { and y} = a\left( \sin t - t \cos t \right),\text { then find the value of } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?
\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] 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 = t2 and y = t3, find \[\frac{d^2 y}{d x^2}\] ?
If y = |x − x2|, then find \[\frac{d^2 y}{d x^2}\] ?
If x = a cos nt − b sin nt, then \[\frac{d^2 x}{d t^2}\] is
If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .
Find the minimum value of (ax + by), where xy = c2.
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]
