1)When v=(2t2+3)4=f(g(t)), where f(t)=t4,g(t)=2t2+3 .
Since dtdf=4t3 and dtdg=4t , by the chain rule, we obtain that:
a=dtdv=dgdfdtdg=4(2t2+3)3⋅4t=16t(2t2+3)3
2)When v=ln(4t3−1)=f(g(t)), where f(t)=ln(4t3−1),g(t)=(4t3−1) .
Since dtdf=4t3−11 and dtdg=12t2 , by the chain rule, we obtain that:
a=dtdv=dgdfdtdg=4t3−11⋅12t2=4t3−112t2
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