Advertisements
Advertisements
प्रश्न
If y = a + bx2, a, b arbitrary constants, then
पर्याय
\[\frac{d^2 y}{d x^2} = 2xy\]
\[x\frac{d^2 y}{d x^2} = y_1\]
\[x\frac{d^2 y}{d x^2} - \frac{dy}{dx} + y = 0\]
\[x\frac{d^2 y}{d x^2} = 2 xy\]
Advertisements
उत्तर
(b) \[x\frac{d^2 y}{d x^2} = y_1\]
Here,
\[y = a + b x^2 \]
\[ \Rightarrow y_1 = 2bx\]
\[ \Rightarrow y_2 = 2b\]
\[\text { Multiplying by x on both sides we get,} \]
\[ x y_2 = 2bx = y_1 \]
\[ \Rightarrow x\frac{d^2 y}{d x^2} = y_1\]
APPEARS IN
संबंधित प्रश्न
Differentiate the following functions from first principles ecos x.
Differentiate the following function from first principles \[e^\sqrt{\cot x}\] .
Differentiate the following functions from first principles log cosec x ?
Differentiate \[\sqrt{\frac{1 + x}{1 - x}}\] ?
Differentiate \[\sin \left( \frac{1 + x^2}{1 - x^2} \right)\] ?
Differentiate \[\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}\] ?
If \[y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}\] ,show that \[\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
If \[y = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] .prove that \[\frac{dy}{dx} = 1 - y^2\] ?
If \[y = \sqrt{a^2 - x^2}\] prove that \[y\frac{dy}{dx} + x = 0\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right), - \pi < x < \pi\] ?
If \[y = \tan^{- 1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right), \text{find } \frac{dy}{dx}\] ?
Find \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?
If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?
Differentiate \[\left( \sin x \right)^{\log x}\] ?
find \[\frac{dy}{dx}\] \[y = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right) \left( 4x - 1 \right)}}\] ?
Find \[\frac{dy}{dx}\] \[y = \sin x \sin 2x \sin 3x \sin 4x\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x}\] ?
If \[y = \left( \sin x - \cos x \right)^{\sin x - \cos x} , \frac{\pi}{4} < x < \frac{3\pi}{4}, \text{ find} \frac{dy}{dx}\] ?
Find \[\frac{dy}{dx}\] , when \[x = \frac{3 at}{1 + t^2}, \text{ and } y = \frac{3 a t^2}{1 + t^2}\] ?
Find \[\frac{dy}{dx}\] , when \[x = \cos^{- 1} \frac{1}{\sqrt{1 + t^2}} \text{ and y } = \sin^{- 1} \frac{t}{\sqrt{1 + t^2}}, t \in R\] ?
Differentiate \[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cos^{- 1} x, \text { if}\] \[x \in \left( - 1, 0 \right)\] ?
Differentiate \[\tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\] with respect to \[\sec^{- 1} x\] ?
If \[y = \sin^{- 1} x + \cos^{- 1} x\] ,find \[\frac{dy}{dx}\] ?
If \[y = \log_a x, \text{ find } \frac{dy}{dx} \] ?
If \[u = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ and v} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] where \[- 1 < x < 1\], then write the value of \[\frac{du}{dv}\] ?
If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\] _________ .
The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]
If \[f\left( x \right) = \left| x^2 - 9x + 20 \right|\] then `f' (x)` is equal to ____________ .
If \[y = \log \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] __________ .
If y = e−x cos x, show that \[\frac{d^2 y}{d x^2} = 2 e^{- x} \sin x\] ?
If y = 2 sin x + 3 cos x, show that \[\frac{d^2 y}{d x^2} + y = 0\] ?
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
If y = 500 e7x + 600 e−7x, show that \[\frac{d^2 y}{d x^2} = 49y\] ?
If y = cosec−1 x, x >1, then show that \[x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\] ?
If \[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}\] .....to ∞, then write \[\frac{d^2 y}{d x^2}\] in terms of y ?
If \[y = \left| \log_e x \right|\] find\[\frac{d^2 y}{d x^2}\] ?
If x = a (1 + cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = \frac{- 1}{a}at \theta = \frac{\pi}{2}\]
