If x = t and y = 1/t^2, then (d^2 y)/(d x^2) is equal to Option 1: 2/t^2 Option 2: 4/t^2 Option 3: -2/t^2 Option 4: -4/t^2

Answer: 4/t^2. Solution: dx/dt = 1/t, dy/dt = -2/t^3. So dy/dx = -2/t^3 * t = -2/t^2. Then d^2y/dx^2 = d/dt(-2/t^2) * dt/dx = 4/t^3 * t = 4/t^2.

Applied Mathematics

Calculus

Medium

applied

If $x = \log t$ and $y = \frac{1}{t^2}$ , then $\frac{d^2 y}{d x^2}$ is equal to

$\frac{2}{t^2}$

$\frac{4}{t^2}$

$-\frac{2}{t^2}$

$-\frac{4}{t^2}$

✅ Correct Option: 2

$\frac{dx}{dt} = \frac{1}{t}$ , $\frac{dy}{dt} = -\frac{2}{t^3}$ . So $\frac{dy}{dx} = -\frac{2}{t^3} \cdot t = -\frac{2}{t^2}$ . Then $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\frac{2}{t^2}\right) \cdot \frac{dt}{dx} = \frac{4}{t^3} \cdot t = \frac{4}{t^2}$ .

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If $x = at^2$ and $y = a^3 t^3$ , then $\frac{d^2 y}{dx^2}$ will be :

If x=log⁡tx = \log tx=logt and y=1t2y = \frac{1}{t^2}y=t21​, then d2ydx2\frac{d^2 y}{d x^2}dx2d2y​ is equal to

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CUET Mathematics Past Year Questions (Topic-Wise)

If $x = \log t$ and $y = \frac{1}{t^2}$ , then $\frac{d^2 y}{d x^2}$ is equal to