Advertisements
Advertisements
Question
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
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 eax+b.
Differentiate \[e^{\tan 3 x} \] ?
Differentiate \[\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)\] ?
If \[y = x \sin^{- 1} x + \sqrt{1 - x^2}\] ,prove that \[\frac{dy}{dx} = \sin^{- 1} x\] ?
If \[y = \sqrt{a^2 - x^2}\] prove that \[y\frac{dy}{dx} + x = 0\] ?
Differentiate \[\sin^{- 1} \left( 2 x^2 - 1 \right), 0 < x < 1\] ?
Differentiate \[\cos^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
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 \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?
Find \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?
Find \[\frac{dy}{dx}\] in the following case \[x^{2/3} + y^{2/3} = a^{2/3}\] ?
If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
Differentiate \[x^\left( \sin x - \cos x \right) + \frac{x^2 - 1}{x^2 + 1}\] ?
Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?
Find \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?
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 \[\frac{dy}{dx} = - \left\{ \frac{x^x \left( 1 + \log x \right) + y^x \cdot \log y}{x \cdot y^\left( x - 1 \right)} \right\}\] ?
Find \[\frac{dy}{dx}\], When \[x = a \left( \theta + \sin \theta \right) \text{ and } y = a \left( 1 - \cos \theta \right)\] ?
Differentiate log (1 + x2) with respect to tan−1 x ?
If \[f'\left( 1 \right) = 2 \text { and y } = f \left( \log_e x \right), \text { find} \frac{dy}{dx} \text { at }x = e\] ?
Let g (x) be the inverse of an invertible function f (x) which is derivable at x = 3. If f (3) = 9 and `f' (3) = 9`, write the value of `g' (9)`.
If \[- \frac{\pi}{2} < x < 0 \text{ and y } = \tan^{- 1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}, \text{ find } \frac{dy}{dx}\] ?
If \[y = \log_a x, \text{ find } \frac{dy}{dx} \] ?
If \[3 \sin \left( xy \right) + 4 \cos \left( xy \right) = 5, \text { then } \frac{dy}{dx} =\] _____________ .
The derivative of \[\cos^{- 1} \left( 2 x^2 - 1 \right)\] with respect to \[\cos^{- 1} x\] is ___________ .
If \[f\left( x \right) = \left| x - 3 \right| \text { and }g\left( x \right) = fof \left( x \right)\] is equal to __________ .
If \[\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3 \left( x^3 - y^3 \right)\] then \[\frac{dy}{dx}\] is equal to ____________ .
If y = x3 log x, prove that \[\frac{d^4 y}{d x^4} = \frac{6}{x}\] ?
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?
If y log (1 + cos x), prove that \[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \cdot \frac{dy}{dx} = 0\] ?
\[\text { If x } = \cos t + \log \tan\frac{t}{2}, y = \sin t, \text { then find the value of } \frac{d^2 y}{d t^2} \text { and } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
\[\text { If x } = a\left( \cos t + t \sin t \right) \text { and y} = a\left( \sin t - t \cos t \right),\text { then find the value of } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
If x = t2 and y = t3, find \[\frac{d^2 y}{d x^2}\] ?
If x = 2at, y = at2, where a is a constant, then find \[\frac{d^2 y}{d x^2} \text { at }x = \frac{1}{2}\] ?
If x = a cos nt − b sin nt, then \[\frac{d^2 x}{d t^2}\] is
If x = 2 at, y = at2, where a is a constant, then \[\frac{d^2 y}{d x^2} \text { at x } = \frac{1}{2}\] is
If \[y = \frac{ax + b}{x^2 + c}\] then (2xy1 + y)y3 =
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]
