question_answer

Differentiate \[{{\sin }^{2}}(3x+1)\] w.r.t.x.

Answer:

Let y =\[{{\sin }^{2}}(3x+1)\] Here, y is a composite of three functions.       Let \[y={{t}^{2}},\] where \[t=\sin \,\,u\,\,\text{and}\,\,u=3x+1\] Then \[\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{du}\cdot \frac{du}{dx}\]             [by Chain rule] \[=\frac{d}{dt}[{{t}^{2}}]\cdot \frac{d}{du}(\sin \,u)\cdot \frac{d}{dx}(3x+1)\] \[=2t(cos\,u)\cdot 3=6t\,cos\,u=6\sin \,u\,\,\cos \,u\] \[[\because \,\,t=\sin \,u]\] = 3 sin2 u         \[[\because \,\,2\sin \theta \,\cos \theta =2sin2\theta ]\] \[=3\sin 2(3x+1)=3\sin (6x+2)\]    \[[\text{put}\,u=3x+1]\]