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If $y = \cos^3(\sec^2 2t)$, find $\frac{dy}{dt}$. |
$-6 \cos^2(\sec^2 2t) \sin(\sec^2 2t) \sec^2(2t) \tan(2t)$ $-12 \cos^2(\sec^2 2t) \sin(\sec^2 2t) \sec^2(2t) \tan(2t)$ $12 \cos^2(\sec^2 2t) \sin(\sec^2 2t) \sec^2(2t) \tan(2t)$ $-12 \cos^3(\sec^2 2t) \sec^2(2t) \tan(2t)$ |
$-12 \cos^2(\sec^2 2t) \sin(\sec^2 2t) \sec^2(2t) \tan(2t)$ |
The correct answer is Option (2) → $-12 \cos^2(\sec^2 2t) \sin(\sec^2 2t) \sec^2(2t) \tan(2t)$ ## $y = \cos^3(\sec^2 2t)$ Applying chain rule, $\frac{dy}{dt} = 3\cos^2(\sec^2 2t) \times [-\sin(\sec^2 2t)] \times 2\sec 2t \times \sec 2t \tan 2t \times 2$ $= -12 \cos^2(\sec^2 2t) \times \sin(\sec^2 2t) \times \sec^2 2t \times \tan 2t$ |
