Question #6bed4

Question

1637 views around the world

You can reuse this answer
Creative Commons License

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

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