Advertisements
Advertisements
प्रश्न
If \[y = \left| \log_e x \right|\] find\[\frac{d^2 y}{d x^2}\] ?
उत्तर
Here,
\[y = \left| \log_e x \right|\]
\[ =\begin{cases} - \log_e x & \text{ if } 0 < x < 1\ \\ \log_e x & \text { if }x > 1\end{cases}\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = \begin{cases}\frac{- 1}{x} & \text { if } 0 < x < 1\\ \frac{1}{x} & \text { if } x > 1\end{cases}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \begin{cases}\frac{1}{x^2} & \text { if } 0 < x < 1\\\frac{- 1}{x^2} & \text { if } x > 1\end{cases}\]
APPEARS IN
संबंधित प्रश्न
Differentiate \[e^{\sin} \sqrt{x}\] ?
Differentiate \[e^{3 x} \cos 2x\] ?
Differentiate \[\frac{x^2 + 2}{\sqrt{\cos x}}\] ?
Differentiate \[\tan^{- 1} \left( \frac{4x}{1 - 4 x^2} \right), - \frac{1}{2} < x < \frac{1}{2}\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - a}{x + a} \right)\] ?
Find \[\frac{dy}{dx}\] in the following case \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] ?
If \[y \sqrt{1 - x^2} + x \sqrt{1 - y^2} = 1\] ,prove that \[\frac{dy}{dx} = - \sqrt{\frac{1 - y^2}{1 - x^2}}\] ?
If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that \[\frac{dy}{dx} = \frac{y}{x}\] ?
If \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?
Differentiate \[\left( 1 + \cos x \right)^x\] ?
Differentiate \[x^{\sin^{- 1} x}\] ?
Differentiate \[x^{\tan^{- 1} x }\] ?
Differentiate\[\left( x + \frac{1}{x} \right)^x + x^\left( 1 + \frac{1}{x} \right)\] ?
Find \[\frac{dy}{dx}\] \[y = x^{\sin x} + \left( \sin x \right)^x\] ?
If \[x^{13} y^7 = \left( x + y \right)^{20}\] prove that \[\frac{dy}{dx} = \frac{y}{x}\] ?
If \[xy \log \left( x + y \right) = 1\] , prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
If \[xy = e^{x - y} , \text{ find } \frac{dy}{dx}\] ?
Find \[\frac{dy}{dx}\] ,When \[x = a \left( 1 - \cos \theta \right) \text{ and } y = a \left( \theta + \sin \theta \right) \text{ at } \theta = \frac{\pi}{2}\] ?
If \[x = a\sin2t\left( 1 + \cos2t \right) \text { and y } = b\cos2t\left( 1 - \cos2t \right)\] , show that at \[t = \frac{\pi}{4}, \frac{dy}{dx} = \frac{b}{a}\] ?
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\sec^{- 1} \left( \frac{1}{\sqrt{1 - x^2}} \right)\], if \[x \in \left( \frac{1}{\sqrt{2}}, 1 \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text { if }0 < x < 1\] ?
If \[f'\left( x \right) = \sqrt{2 x^2 - 1} \text { and y } = f \left( x^2 \right)\] then find \[\frac{dy}{dx} \text { at } x = 1\] ?
If \[y = x \left| x \right|\] , find \[\frac{dy}{dx} \text{ for } x < 0\] ?
If \[y = \log \sqrt{\tan x}, \text{ write } \frac{dy}{dx} \] ?
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
If \[\sin y = x \cos \left( a + y \right), \text { then } \frac{dy}{dx}\] is equal to ______________ .
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 = tan−1 x, show that \[\left( 1 + x^2 \right) \frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\] ?
Find \[\frac{d^2 y}{d x^2}\] where \[y = \log \left( \frac{x^2}{e^2} \right)\] ?
\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?
\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] 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 y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
If \[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\] then (1 − x)2 f '' (x) − xf(x) =
If xy − loge y = 1 satisfies the equation \[x\left( y y_2 + y_1^2 \right) - y_2 + \lambda y y_1 = 0\]
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]