Question #6bed4

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Answer 1 · 2017-01-23T01:44:51

#(-6t^2sin(t^3))'=-12tsin(t^3)-18t^4cos(t^3)#

First remember the product rule

#(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)#

In this case

#f(x)=-6t^2#

and

#g(x)=sin(t^3)#

then

since

#(cx^n)'=c(x^n)'=cnx^(n-1)# power rule and derivative times a constant equals constant times derivative

#f'(x)=-12t#

also by the chain rule #(f(g(x)))'=f'(g(x))g'(x)#

#g'(x)=3t^2cos(t^3)#

Then

#(-6t^2sin(t^3))'=-12tsin(t^3)-18t^4cos(t^3)#