The second derivative of a single valued function parametrically represented by x - t and y - t - where - t and - t are different functions and - t -0 is given byA- d2 y-d x2-d x-d td2 y-d t2-d2 x-d t2d y-d t-d x-d t3B- d2 y-d x2-d x-d td2 y-d t2-d2 x-d t2d y-d t-d x-d t2C- d2 y-d x2-d2 x-d t2 d y-d t-d x-d td2 y-d t2-d x3-d tD- d2 y-d x2-d2 x-d t2d y-d t-d2 y-d t2 d x-d t-d y-d y3

The correct option is A dy/dx=dy/dt/dx/dt d^2y/dx^2=d/dx(dy/dt/dx/dt) ...

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Differentiability in an Interval

The second de...

Question

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

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x = f (t), y = g (t)

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

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

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

By introducing a new variable t, putting x=cos t , the expression is $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y$ transformed into :

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

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