D 2 x - dv 2 equalsA- -d2 y-d x2-1d y-d x-3B- d2 y-d x2d y-d x-2C- -d2 y-d x2-d y-d x-3D- d2 y-d x2-1

The correct option is C d^2x/dy^2=d/dy(d/dx )=d/dx(dx/dy )×dx/dy = d/dx [1/dy/dx ]×1/(dy/dx ) = 1/ (dy/dx )^2×d^2y/dx^2×1/dy/dx = (dy/dx )^ 3.d^2y/dx^2 ...

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Byju's Answer

Standard XI

Mathematics

Proof by mathematical induction

d 2 x / dv 2 ...

Question

$\frac{d^{2} x}{d y^{2}} e q u a l s$

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Solution

The correct option is D
$\frac{d^{2} x}{d y^{2}} = \frac{d}{d y} (\frac{d}{d x}) = \frac{d}{d x} (\frac{d x}{d y}) \times \frac{d x}{d y} = \frac{d}{d x} [\frac{\frac{1}{d y}}{d x}] \times \frac{1}{(\frac{d y}{d x})} = \frac{- 1}{{(\frac{d y}{d x})}^{2}} \times \frac{d^{2} y}{d x^{2}} \times \frac{\frac{1}{d y}}{d x} = - {(\frac{d y}{d x})}^{- 3} . \frac{d^{2} y}{d x^{2}}$

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Similar questions

$\frac{d^{2} x}{d y^{2}}$ equals:

\frac{d^{2} x}{d y^{2}} e q u a l s

\frac{d^{2} x}{d y^{2}}

equals:

The second derivative of a single valued function parametrically represented by $x = ϕ (t) a n d y = ψ (t)$ , ( where $ϕ (t) a n d ψ (t)$ are different functions and $ϕ^{'} (t) \neq 0$ ) is given by

\frac{d^{2} x}{d y^{2}} = - \frac{d^{2} y}{d x^{2}} . {(\frac{d x}{d y})}^{3}

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d2xdy2 equals

The correct option is D d2xdy2=ddy(ddx)=ddx(dxdy)×dxdy=ddx[1dydx]×1(dydx)=−1(dydx)2×d2ydx2×1dydx=−(dydx)−3.d2ydx2

$\frac{d^{2} x}{d y^{2}} e q u a l s$