Advertisements
Advertisements
प्रश्न
Find \[\frac{dy}{dx}\] in the following case \[4x + 3y = \log \left( 4x - 3y \right)\] ?
Advertisements
उत्तर
We have,
\[4x + 3y = \log\left( 4x - 3y \right)\]
Differentiating with respect to x, we get,
\[\frac{d}{dx}\left( 4x \right) + \frac{d}{dx}\left( 3y \right) = \frac{d}{dx}\left\{ \log\left( 4x - 3y \right) \right\}\]
\[ \Rightarrow 4 + 3\frac{d y}{d x} = \frac{1}{\left( 4x - 3y \right)}\frac{d}{dx}\left( 4x - 3y \right) \]
\[ \Rightarrow 4 + 3\frac{d y}{d x} = \frac{1}{\left( 4x - 3y \right)}\left( 4 - 3\frac{d y}{d x} \right)\]
\[ \Rightarrow 3\frac{d y}{d x} + \frac{3}{\left( 4x - 3y \right)}\frac{d y}{d x} = \frac{4}{\left( 4x - 3y \right)} - 4\]
\[ \Rightarrow 3\frac{d y}{d x}\left\{ 1 + \frac{1}{\left( 4x - 3y \right)} \right\} = 4\left\{ \frac{1}{\left( 4x - 3y \right)} - 1 \right\}\]
\[ \Rightarrow 3\frac{d y}{d x}\left\{ \frac{4x - 3y + 1}{\left( 4x - 3y \right)} \right\} = 4\left\{ \frac{1 - 4x + 3y}{\left( 4x - 3y \right)} \right\}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{4}{3}\left\{ \frac{1 - 4x + 3y}{\left( 4x - 3y \right)} \right\}\left( \frac{4x - 3y}{4x - 3y + 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{4}{3}\left( \frac{1 - 4x + 3y}{4x - 3y + 1} \right)\]
APPEARS IN
संबंधित प्रश्न
If the function f(x)=2x3−9mx2+12m2x+1, where m>0 attains its maximum and minimum at p and q respectively such that p2=q, then find the value of m.
Differentiate \[\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}\] ?
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Differentiate \[\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}\] ?
If \[y = e^x \cos x\] ,prove that \[\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)\] ?
If \[y = \sqrt{a^2 - x^2}\] prove that \[y\frac{dy}{dx} + x = 0\] ?
Differentiate \[\sin^{- 1} \left\{ \sqrt{1 - x^2} \right\}, 0 < x < 1\] ?
Differentiate
\[\tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right), \frac{\pi}{4} < x < \frac{\pi}{4}\] ?
If \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?
Find \[\frac{dy}{dx}\] in the following case \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] ?
Differentiate \[\left( 1 + \cos x \right)^x\] ?
Differentiate \[\left( \log x \right)^{\cos x}\] ?
Differentiate \[x^\left( \sin x - \cos x \right) + \frac{x^2 - 1}{x^2 + 1}\] ?
Differentiate \[\left( \cos x \right)^x + \left( \sin x \right)^{1/x}\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
Find \[\frac{dy}{dx}\]
\[y = x^x + x^{1/x}\] ?
If \[x^m y^n = 1\] , prove that \[\frac{dy}{dx} = - \frac{my}{nx}\] ?
If \[e^y = y^x ,\] prove that\[\frac{dy}{dx} = \frac{\left( \log y \right)^2}{\log y - 1}\] ?
If \[e^{x + y} - x = 0\] ,prove that \[\frac{dy}{dx} = \frac{1 - x}{x}\] ?
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 \[x = \frac{2 t}{1 + t^2} \text{ and } y = \frac{1 - t^2}{1 + t^2}\] ?
If \[x = 10 \left( t - \sin t \right), y = 12 \left( 1 - \cos t \right), \text { find } \frac{dy}{dx} .\] ?
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\sec^{- 1} \left( \frac{1}{\sqrt{1 - x^2}} \right)\], if \[x \in \left( 0, \frac{1}{\sqrt{2}} \right)\] ?
If f (x) = loge (loge x), then write the value of `f' (e)` ?
Given \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .
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 _____________ .
Find the second order derivatives of the following function sin (log x) ?
If y = log (sin x), prove that \[\frac{d^3 y}{d x^3} = 2 \cos \ x \ {cosec}^3 x\] ?
If x = a sec θ, y = b tan θ, prove that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
If x = cos θ, y = sin3 θ, prove that \[y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 = 3 \sin^2 \theta\left( 5 \cos^2 \theta - 1 \right)\] ?
If y = ae2x + be−x, show that, \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\] ?
If x = t2 and y = t3, find \[\frac{d^2 y}{d x^2}\] ?
If \[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\] then (1 − x)2 f '' (x) − xf(x) =
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 \[\frac{d^2 y}{d x^2} =\]
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 =
Find the minimum value of (ax + by), where xy = c2.
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]
