CUET UG 2025 — Mathematics Statistics & Applications

Given $x = e^{\cos 2t}$x=ecos2t and $y = e^{\sin 2t}$y=esin2t.

When both variables depend on a parameter $t$t:

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

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For $x = e^{\cos 2t}$x=ecos2t:

$$\frac{dx}{dt} = e^{\cos 2t} \times \frac{d}{dt}(\cos 2t)$$dtdx​=ecos2t×dtd​(cos2t)

$$\frac{dx}{dt} = e^{\cos 2t} \times (-\sin 2t) \times 2$$dtdx​=ecos2t×(−sin2t)×2

$$\frac{dx}{dt} = -2\sin 2t \cdot e^{\cos 2t}$$dtdx​=−2sin2t⋅ecos2t

$$\frac{dx}{dt} = -2x\sin 2t$$dtdx​=−2xsin2t

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For $y = e^{\sin 2t}$y=esin2t:

$$\frac{dy}{dt} = e^{\sin 2t} \times \frac{d}{dt}(\sin 2t)$$dtdy​=esin2t×dtd​(sin2t)

$$\frac{dy}{dt} = e^{\sin 2t} \times \cos 2t \times 2$$dtdy​=esin2t×cos2t×2

$$\frac{dy}{dt} = 2y\cos 2t$$dtdy​=2ycos2t

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$$\frac{dy}{dx} = \frac{2y\cos 2t}{-2x\sin 2t}$$dxdy​=−2xsin2t2ycos2t​

$$\frac{dy}{dx} = \frac{-y\cos 2t}{x\sin 2t}$$dxdy​=xsin2t−ycos2t​

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From $x = e^{\cos 2t}$x=ecos2t:

$$\ln x = \cos 2t$$lnx=cos2t

$$\cos 2t = \log_e x$$cos2t=loge​x

From $y = e^{\sin 2t}$y=esin2t:

$$\ln y = \sin 2t$$lny=sin2t

$$\sin 2t = \log_e y$$sin2t=loge​y

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$$\frac{dy}{dx} = \frac{-y\cos 2t}{x\sin 2t}$$dxdy​=xsin2t−ycos2t​

$$\frac{dy}{dx} = \frac{-y \cdot \log_e x}{x \cdot \log_e y}$$dxdy​=x⋅loge​y−y⋅loge​x​

Therefore, $\frac{dy}{dx} = \frac{-y\log_e x}{x\log_e y}$dxdy​=xloge​y−yloge​x​