Advertisements
Advertisements
प्रश्न
If \[e^x + e^y = e^{x + y}\] , prove that
\[\frac{dy}{dx} + e^{y - x} = 0\] ?
उत्तर
\[\text{ We have}, e^x + e^y = e^{x + y}\]
Differentiating both sides using chain rule,
\[\frac{d}{dx}\left( e^x \right) + \frac{d}{dx}\left( e^y \right) = \frac{d}{dx}\left( e^{x + y} \right)\]
\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} \frac{d}{dx}\left( x + y \right)\]
\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} \left[ 1 + \frac{dy}{dx} \right]\]
\[ \Rightarrow e^y \frac{dy}{dx} - e^{x + y} \frac{dy}{dx} = e^{x + y} - e^x \]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^{x + y} - e^x}{e^y - e^{x + y}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^{x + y} - e^{x + y} + e^y}{e^{x + y} - e^x - e^{x + y}} \left[ \text{Using eqn } . \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^y}{- e^x}\]
\[ \Rightarrow \frac{dy}{dx} = - e^{y - x} \]
\[ \Rightarrow \frac{dy}{dx} + e^{y - x} = 0\]
APPEARS IN
संबंधित प्रश्न
Differentiate the following functions from first principles log cosec x ?
Differentiate \[\log \left( x + \sqrt{x^2 + 1} \right)\] ?
Differentiate \[\sin \left( 2 \sin^{- 1} x \right)\] ?
Differentiate \[\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}\] ?
Differentiate \[\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}\] ?
\[\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\] ?
Differentiate \[\cos \left( \log x \right)^2\] ?
If \[y = \log \sqrt{\frac{1 + \tan x}{1 - \tan x}}\] prove that \[\frac{dy}{dx} = \sec 2x\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
Differentiate \[\cos^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?
Differentiate \[\tan^{- 1} \left( \frac{2 a^x}{1 - a^{2x}} \right), a > 1, - \infty < x < 0\] ?
Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] ?
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 \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
If \[y = \cos^{- 1} \left\{ \frac{2x - 3 \sqrt{1 - x^2}}{\sqrt{13}} \right\}, \text{ find } \frac{dy}{dx}\] ?
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 \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
Differentiate \[x^{1/x}\] with respect to x.
Differentiate \[{10}^{ \log \sin x }\] ?
Differentiate \[\left( \log x \right)^{ \log x }\] ?
Differentiate \[\left( \cos x \right)^x + \left( \sin x \right)^{1/x}\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
Find \[\frac{dy}{dx}\] \[y = x^{\sin x} + \left( \sin x \right)^x\] ?
Find \[\frac{dy}{dx}\] \[y = x^{\cos x} + \left( \sin x \right)^{\tan x}\] ?
If \[y = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}}\], prove that \[\frac{dy}{dx} = 2\ at\ x = \frac{\pi}{4}\] ?
Differentiate log (1 + x2) with respect to tan−1 x ?
Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2}, - \frac{1}{2 \sqrt{2}} \right)\] ?
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 \[- 1 < x < 1, x \neq 0 .\] ?
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right), \text { if }- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?
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 \[\sin y = x \cos \left( a + y \right), \text { then } \frac{dy}{dx}\] is equal to ______________ .
If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then } \frac{dy}{dx}\] is equal to ___________ .
If y = e−x cos x, show that \[\frac{d^2 y}{d x^2} = 2 e^{- x} \sin x\] ?
If \[y = \frac{\log x}{x}\] show that \[\frac{d^2 y}{d x^2} = \frac{2 \log x - 3}{x^3}\] ?
\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
If y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
If x = a (1 + cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = \frac{- 1}{a}at \theta = \frac{\pi}{2}\]
f(x) = 3x2 + 6x + 8, x ∈ R
