If x - 1 - t1 - t - y - 2t1 - t then d2ydx2 -

The correct option is A 0 dydt = (1 + t)2 2t(1 + t)^2 = 2(1 + #10; t)^2 | dxdt = (1 + t) (1 t)(1 + t)^2 = #10; 2(1 + t)^2 ∴dydx = 1 ...

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Higher Order Derivatives

If x = 1 - ...

Question

If $x = \frac{1 - t}{1 + t} : y = \frac{2 t}{1 + t}$ then $\frac{d^{2} y}{d x^{2}} =$

\frac{2 t}{(1 + t)^{2}}

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\frac{1}{(1 + t)^{4}}

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\frac{2 t^{2}}{(1 + t)^{2}}

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Solution

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Similar questions

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

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

t = 0

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

t

cos x = \frac{1 - t^{2}}{1 + t^{2}}

sin y = \frac{2 t}{1 + t^{2}}

t \in (- 1, 0)

4 \frac{d^{2} y}{d x^{2}} - \frac{3}{2} \frac{d y}{d x} - \frac{1}{2} \frac{y}{x}

(x, y) = (1, - 1)

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

\frac{d y}{d x}

t = 2

| t | < 1

sin x = \frac{2 t}{1 + t^{2}}

tan y = \frac{2 t}{1 - t^{2}}

\frac{d y}{d x}

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