![\dfrac{\mathrm d}{\mathrm dt}[\cos2t]=-\sin2t\cdot\dfrac{\mathrm d}{\mathrm dt}[2t]](/tpl/images/0499/5259/4e7ad.png)
![\dfrac{\mathrm d}{\mathrm dt}[\sin t^2]=\cos t^2\cdot\dfrac{\mathrm d}{\mathrm dt}[t^2}](/tpl/images/0499/5259/e959b.png)
![\dfrac{\mathrm d}{\mathrm dt}[\sin^2t]=2\sin t\cdot\dfrac{\mathrm d}{\mathrm dt}[\sin t]](/tpl/images/0499/5259/afb79.png)
, as opposed to 
.![\dfrac{\mathrm d}{\mathrm dt}[\sin(2t+1)^2]=\cos(2t+1)^2\cdot\dfrac{\mathrm d}{\mathrm dt}[(2t+1)^2]](/tpl/images/0499/5259/13ec5.png)
![=2(2t+1)\cos(2t+1)^2\cdot\dfrac{\mathrm d}{\mathrm dt}[2t+1]](/tpl/images/0499/5259/3ff93.png)
Iwould like some with those 4 . calculus! show your work ! !
...
Mathematics, 04.02.2020 06:54 linaaaaa7
Iwould like some with those 4 . calculus! show your work ! !
Answers: 1
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