Question

Find \(\frac{dy}{dx}\), if x=sin⁡3t and y=t^2 tan⁡2t.

(a) \(\frac{3t(tan⁡2t+tsec^2 2t)}{4 cos⁡3t}\)

(b) \(\frac{(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)

(c) \(\frac{-2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)

(d) \(\frac{2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)

I have been asked this question in examination.

My question is from Derivatives of Functions in Parametric Forms topic in chapter Continuity and Differentiability of Mathematics – Class 12

Answer 1

Right option is (d) \(\frac{2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)

To explain I would say: Given that, x=sin⁡3t and y=t^2 tan⁡2t

\(\frac{dx}{dt}\)=3 cos⁡3t

By using u.v rule, we get

\(\frac{dy}{dt}\)=\(\frac{d}{dx} \,(t^2) \,tan⁡2t+\frac{d}{dx} \,(tan⁡2t)\) t^2

\(\frac{dy}{dt}\)=2t tan⁡2t+2t^2 sec^2⁡2t

\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{(2t \,tan⁡2t+2t^2 \,sec^2⁡2t)}{3 \,cos⁡3t}\)

∴\(\frac{dy}{dx}=\frac{2t(tan⁡2t+tsec^2 \,2t)}{3 \,cos⁡3t}\)