Advertisements
Advertisements
Question
If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is
Options
a constant
a function of x only
a function of y only
a function of x and y
Advertisements
Solution
(a) a constant
Here,
\[y^2 = a x^2 + bx + c\]
\[\text { Now,} \]
\[2y\frac{d y}{d x} = 2ax + b\]
\[ \Rightarrow 2y\frac{d^2 y}{d x^2} + 2 \left( \frac{d y}{d x} \right)^2 = 2a \]
\[ \Rightarrow y\frac{d^2 y}{d x^2} + \left( \frac{d y}{d x} \right)^2 = a \]
\[ \Rightarrow y\frac{d^2 y}{d x^2} + \left( \frac{2ax + b}{2y} \right)^2 = a \left[ \because 2y\frac{d y}{d x} = 2ax + b \right]\]
\[ \Rightarrow 4 y^3 \frac{d^2 y}{d x^2} + \left( 2ax + b \right)^2 = 4a y^2 \]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4a y^2 - \left( 2ax + b \right)^2}{4}\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4a\left( a x^2 + bx + c \right) - \left( 2ax + b \right)^2}{4} \left[ \because y^2 = a x^2 + bx + c \right]\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4 a^2 x^2 + 4abx + 4ac - 4 a^2 x^2 - b^2 - 4axb}{4}\]
\[ \Rightarrow y^3 \frac{d^2 y}{d x^2} = \frac{4ac - b^2}{4} = \text { a constant }\]
APPEARS IN
RELATED QUESTIONS
Prove that `y=(4sintheta)/(2+costheta)-theta `
Differentiate the following functions from first principles x2ex ?
Differentiate the following functions from first principles sin−1 (2x + 3) ?
Differentiate tan (x° + 45°) ?
Differentiate \[3^{x^2 + 2x}\] ?
Differentiate \[\log \left( \frac{\sin x}{1 + \cos x} \right)\] ?
Differentiate \[\tan^{- 1} \left( e^x \right)\] ?
If \[y = \sqrt{x^2 + a^2}\] prove that \[y\frac{dy}{dx} - x = 0\] ?
If \[y = \sqrt{a^2 - x^2}\] prove that \[y\frac{dy}{dx} + x = 0\] ?
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
Differentiate \[\sin^{- 1} \left( 1 - 2 x^2 \right), 0 < x < 1\] ?
Differentiate \[\tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}, - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - a}{x + a} \right)\] ?
Find \[\frac{dy}{dx}\] in the following case \[xy = c^2\] ?
If \[x y^2 = 1,\] prove that \[2\frac{dy}{dx} + y^3 = 0\] ?
If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find} \frac{dy}{dx}\] ?
If \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find} \frac{dy}{dx}\] ?
If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?
Differentiate \[x^{\cos^{- 1} x}\] ?
If \[xy = e^{x - y} , \text{ find } \frac{dy}{dx}\] ?
If \[y = \left( \cos x \right)^{\left( \cos x \right)^{\left( \cos x \right) . . . \infty}}\],prove that \[\frac{dy}{dx} = - \frac{y^2 \tan x}{\left( 1 - y \log \cos x \right)}\]?
Find \[\frac{dy}{dx}\] when \[x = \frac{2 t}{1 + t^2} \text{ and } y = \frac{1 - t^2}{1 + t^2}\] ?
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 \[y = \sin^{- 1} \left( \sin x \right), - \frac{\pi}{2} \leq x \leq \frac{\pi}{2}\] ,Then, write the value of \[\frac{dy}{dx} \text{ for } x \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \] ?
If \[f\left( 0 \right) = f\left( 1 \right) = 0, f'\left( 1 \right) = 2 \text { and y } = f \left( e^x \right) e^{f \left( x \right)}\] write the value of \[\frac{dy}{dx} \text{ at x } = 0\] ?
If f (x) = logx2 (log x), the `f' (x)` at x = e is ____________ .
If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0 ?
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
If y = ae2x + be−x, show that, \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\] ?
If y = ex (sin x + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?
If y log (1 + cos x), prove that \[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \cdot \frac{dy}{dx} = 0\] ?
If x = f(t) and y = g(t), then write the value of \[\frac{d^2 y}{d x^2}\] ?
Let f(x) be a polynomial. Then, the second order derivative of f(ex) is
If x = f(t) and y = g(t), then \[\frac{d^2 y}{d x^2}\] is equal to
If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .
