CUET UG 2025 — Mathematics Calculus

Given $y = t - \frac{1}{t}$y=t−t1​ and $x = t + \frac{1}{t}$x=t+t1​, where both x and y depend on parameter t.

When both variables are given in terms of a parameter, the derivative is:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$dxdy​=dx/dtdy/dt​

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Differentiating y with respect to t:

$$y = t - \frac{1}{t}$$y=t−t1​

$$\frac{dy}{dt} = 1 - \left(-\frac{1}{t^2}\right)$$dtdy​=1−(−t21​)

$$\frac{dy}{dt} = 1 + \frac{1}{t^2}$$dtdy​=1+t21​

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Differentiating x with respect to t:

$$x = t + \frac{1}{t}$$x=t+t1​

$$\frac{dx}{dt} = 1 + \left(-\frac{1}{t^2}\right)$$dtdx​=1+(−t21​)

$$\frac{dx}{dt} = 1 - \frac{1}{t^2}$$dtdx​=1−t21​

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Applying the formula:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$dxdy​=dx/dtdy/dt​

$$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}$$dxdy​=1−t21​1+t21​​

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Multiplying numerator and denominator by $t^2$t2:

$$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \times \frac{t^2}{t^2}$$dxdy​=1−t21​1+t21​​×t2t2​

$$\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$$dxdy​=t2−1t2+1​

Therefore, $\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$dxdy​=t2−1t2+1​